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completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-16x+76$ $f(x)=(x+$ | We want to complete $x^2{-16}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-16}}{2}\right)^2={64}$ to it: $x^2{-16}x+{64}=(x-8)^2$ In order to keep the expression equivalent, we add and subtract ${64}$, not forgetting the expression's constant term, $76$ : $\begin{aligned} f(x)&=x^2-16x+76 \\\\ &=x^2-16x+{64}+76-{64} \\\\ &=(x-8)^2+76-64 \\\\ &=(x-8)^2+12 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 8)^2 +12$ This is equivalent to $f(x)=(x+{-8})^2+12$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-14x+33 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 - 14x = -33$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-14$, half of it would be $-7$, and squaring it gives us ${49}$. $x^2 - 14x { + 49} = -33 { + 49}$ We can now rewrite the left side of the equation as a squared term. $( x - 7 )^2 = 16$ This is equivalent to $(x+{-7})^2=16$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+6x+9 = 0$ $(x + $ | The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $6$, half of it is $3$, and squaring it gives us ${9}$, our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x + 3 )^2 = 0$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-20x+100 = 0$ $(x + $ | The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $-20$, half of it is $-10$, and squaring it gives us ${100}$, our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x - 10 )^2 = 0$ This is equivalent to $(x+{-10})^2=0$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-6x+43$ $f(x)=(x+$ | We want to complete $x^2{-6}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-6}}{2}\right)^2={9}$ to it: $x^2{-6}x+{9}=(x-3)^2$ In order to keep the expression equivalent, we add and subtract ${9}$, not forgetting the expression's constant term, $43$ : $\begin{aligned} f(x)&=x^2-6x+43 \\\\ &=x^2-6x+{9}+43-{9} \\\\ &=(x-3)^2+43-9 \\\\ &=(x-3)^2+34 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 3)^2 + 34$ This is equivalent to $f(x)=(x+{-3})^2+34$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-16x+63 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 - 16x = -63$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-16$, half of it would be $-8$, and squaring it gives us ${64}$. $x^2 - 16x { + 64} = -63 { + 64}$ We can now rewrite the left side of the equation as a squared term. $( x - 8 )^2 = 1$ This is equivalent to $(x+{-8})^2=1$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-2x-35 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 - 2x = 35$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-2$, half of it would be $-1$, and squaring it gives us ${1}$. $x^2 - 2x { + 1} = 35 { + 1}$ We can now rewrite the left side of the equation as a squared term. $( x - 1 )^2 = 36$ This is equivalent to $(x+{-1})^2=36$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+8x-29$ $f(x)=(x+$ | We want to complete $x^2{+8}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+8}}{2}\right)^2={16}$ to it: $x^2{+8}x+{16}=(x+4)^2$ In order to keep the expression equivalent, we add and subtract ${16}$, not forgetting the expression's constant term, $-29$ : $\begin{aligned} f(x)&=x^2+8x-29 \\\\ &=x^2+8x+{16}-29-{16} \\\\ &=(x+4)^2-29-16 \\\\ &=(x+4)^2-45 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 4)^2 - 45$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+10x+21 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 10x = -21$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $10$, half of it would be $5$, and squaring it gives us ${25}$. $x^2 + 10x { + 25} = -21 { + 25}$ We can now rewrite the left side of the equation as a squared term. $( x + 5 )^2 = 4$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+14x+64$ $f(x)=(x+$ | We want to complete $x^2{+14}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+14}}{2}\right)^2={49}$ to it: $x^2{+14}x+{49}=(x+7)^2$ In order to keep the expression equivalent, we add and subtract ${49}$, not forgetting the expression's constant term, $64$ : $\begin{aligned} f(x)&=x^2+14x+64 \\\\ &=x^2+14x+{49}+64-{49} \\\\ &=(x+7)^2+64-49 \\\\ &=(x+7)^2+15 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 7)^2 +15$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+14x+8$ $f(x)=(x+$ | We want to complete $x^2{+14}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+14}}{2}\right)^2={49}$ to it: $x^2{+14}x+{49}=(x+7)^2$ In order to keep the expression equivalent, we add and subtract ${49}$, not forgetting the expression's constant term, $8$ : $\begin{aligned} f(x)&=x^2+14x+8 \\\\ &=x^2+14x+{49}+8-{49} \\\\ &=(x+7)^2+8-49 \\\\ &=(x+7)^2-41 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 7)^2 - 41$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+10x+25 = 0$ $(x + $ | The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $10$, half of it is $5$, and squaring it gives us ${25}$, our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x + 5 )^2 = 0$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-8x-51$ $f(x)=(x+$ | We want to complete $x^2{-8}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-8}}{2}\right)^2={16}$ to it: $x^2{-8}x+{16}=(x-4)^2$ In order to keep the expression equivalent, we add and subtract ${16}$, not forgetting the expression's constant term, $-51$ : $\begin{aligned} f(x)&=x^2-8x-51 \\\\ &=x^2-8x+{16}-51-{16} \\\\ &=(x-4)^2-51-16 \\\\ &=(x-4)^2-67 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 4)^2 - 67$ This is equivalent to $f(x)=(x+{-4})^2-67$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+20x+40$ $f(x)=(x+$ | We want to complete $x^2{+20}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+20}}{2}\right)^2={100}$ to it: $x^2{+20}x+{100}=(x+10)^2$ In order to keep the expression equivalent, we add and subtract ${100}$, not forgetting the expression's constant term, $40$ : $\begin{aligned} f(x)&=x^2+20x+40 \\\\ &=x^2+20x+{100}+40-{100} \\\\ &=(x+10)^2+40-100 \\\\ &=(x+10)^2-60 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 10)^2 - 60$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-16x-100$ $f(x)=(x+$ | We want to complete $x^2{-16}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-16}}{2}\right)^2={64}$ to it: $x^2{-16}x+{64}=(x-8)^2$ In order to keep the expression equivalent, we add and subtract ${64}$, not forgetting the expression's constant term, $-100$ : $\begin{aligned} f(x)&=x^2-16x-100 \\\\ &=x^2-16x+{64}-100-{64} \\\\ &=(x-8)^2-100-64 \\\\ &=(x-8)^2-164 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 8)^2 -164$ This is equivalent to $f(x)=(x+{-8})^2-164$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+20x-86$ $f(x)=(x+$ | We want to complete $x^2{+20}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+20}}{2}\right)^2={100}$ to it: $x^2{+20}x+{100}=(x+10)^2$ In order to keep the expression equivalent, we add and subtract ${100}$, not forgetting the expression's constant term, $-86$ : $\begin{aligned} f(x)&=x^2+20x-86 \\\\ &=x^2+20x+{100}-86-{100} \\\\ &=(x+10)^2-86-100 \\\\ &=(x+10)^2-186 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 10)^2 - 186$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+2x-8 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 2x = 8$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $2$, half of it would be $1$, and squaring it gives us ${1}$. $x^2 + 2x { + 1} = 8 { + 1}$ We can now rewrite the left side of the equation as a squared term. $( x + 1 )^2 = 9$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-10x+44$ $f(x)=(x+$ | We want to complete $x^2{-10}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-10}}{2}\right)^2={25}$ to it: $x^2{-10}x+{25}=(x-5)^2$ In order to keep the expression equivalent, we add and subtract ${25}$, not forgetting the expression's constant term, $44$ : $\begin{aligned} f(x)&=x^2-10x+44 \\\\ &=x^2-10x+{25}+44-{25} \\\\ &=(x-5)^2+44-25 \\\\ &=(x-5)^2+19 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 5)^2 +19$ This is equivalent to $f(x)=(x+{-5})^2+19$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+16x+64 = 0$ $(x + $ | The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $16$, half of it is $8$, and squaring it gives us ${64}$, our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x + 8 )^2 = 0$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+2x-48 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 2x = 48$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $2$, half of it would be $1$, and squaring it gives us ${1}$. $x^2 + 2x { + 1} = 48 { + 1}$ We can now rewrite the left side of the equation as a squared term. $( x + 1 )^2 = 49$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+12x-69$ $f(x)=(x+$ | We want to complete $x^2{+12}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+12}}{2}\right)^2={36}$ to it: $x^2{+12}x+{36}=(x+6)^2$ In order to keep the expression equivalent, we add and subtract ${36}$, not forgetting the expression's constant term, $-69$ : $\begin{aligned} f(x)&=x^2+12x-69 \\\\ &=x^2+12x+{36}-69-{36} \\\\ &=(x+6)^2-69-36 \\\\ &=(x+6)^2-105 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 6)^2 - 105$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-4x-26$ $f(x)=(x+$ | We want to complete $x^2{-4}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-4}}{2}\right)^2={4}$ to it: $x^2{-4}x+{4}=(x-2)^2$ In order to keep the expression equivalent, we add and subtract ${4}$, not forgetting the expression's constant term, $-26$ : $\begin{aligned} f(x)&=x^2-4x-26 \\\\ &=x^2-4x+{4}-26-{4} \\\\ &=(x-2)^2-26-4 \\\\ &=(x-2)^2-30 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 2)^2 -30$ This is equivalent to $f(x)=(x+{-2})^2-30$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+16x-46$ $f(x)=(x+$ | We want to complete $x^2{+16}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+16}}{2}\right)^2={64}$ to it: $x^2{+16}x+{64}=(x+8)^2$ In order to keep the expression equivalent, we add and subtract ${64}$, not forgetting the expression's constant term, $-46$ : $\begin{aligned} f(x)&=x^2+16x-46 \\\\ &=x^2+16x+{64}-46-{64} \\\\ &=(x+8)^2-46-64 \\\\ &=(x+8)^2-110 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 8)^2 - 110$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+2x+26$ $f(x)=(x+$ | We want to complete $x^2{+2}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+2}}{2}\right)^2={1}$ to it: $x^2{+2}x+{1}=(x+1)^2$ In order to keep the expression equivalent, we add and subtract ${1}$, not forgetting the expression's constant term, $26$ : $\begin{aligned} f(x)&=x^2+2x+26 \\\\ &=x^2+2x+{1}+26-{1} \\\\ &=(x+1)^2+26-1 \\\\ &=(x+1)^2+25 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 1)^2 + 25$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+10x+24 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 10x = -24$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $10$, half of it would be $5$, and squaring it gives us ${25}$. $x^2 + 10x { + 25} = -24 { + 25}$ We can now rewrite the left side of the equation as a squared term. $( x + 5 )^2 = 1$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-2x-95$ $f(x)=(x+$ | We want to complete $x^2{-2}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-2}}{2}\right)^2={1}$ to it: $x^2{-2}x+{1}=(x-1)^2$ In order to keep the expression equivalent, we add and subtract ${1}$, not forgetting the expression's constant term, $-95$ : $\begin{aligned} f(x)&=x^2-2x-95 \\\\ &=x^2-2x+{1}-95-{1} \\\\ &=(x-1)^2-95-1 \\\\ &=(x-1)^2-96 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 1)^2 - 96$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-2x+1 = 0$ $(x + $ | The left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $-2$, half of it is $-1$, and squaring it gives us ${1}$, our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x - 1 )^2 = 0$ This is equivalent to $(x+{-1})^2=0$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+18x+80 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 18x = -80$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $18$, half of it would be $9$, and squaring it gives us ${81}$. $x^2 + 18x { + 81} = -80 { + 81}$ We can now rewrite the left side of the equation as a squared term. $( x + 9 )^2 = 1$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-6x-16 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 - 6x = 16$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-6$, half of it would be $-3$, and squaring it gives us ${9}$. $x^2 - 6x { + 9} = 16 { + 9}$ We can now rewrite the left side of the equation as a squared term. $( x - 3 )^2 = 25$ This is equivalent to $(x+{-3})^2=25$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+6x+87$ $f(x)=(x+$ | We want to complete $x^2{+6}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+6}}{2}\right)^2={9}$ to it: $x^2{+6}x+{9}=(x+3)^2$ In order to keep the expression equivalent, we add and subtract ${9}$, not forgetting the expression's constant term, $87$ : $\begin{aligned} f(x)&=x^2+6x+87 \\\\ &=x^2+6x+{9}+87-{9} \\\\ &=(x+3)^2+87-9 \\\\ &=(x+3)^2+78 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 3)^2 + 78$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+2x-39$ $f(x)=(x+$ | We want to complete $x^2{+2}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+2}}{2}\right)^2={1}$ to it: $x^2{+2}x+{1}=(x+1)^2$ In order to keep the expression equivalent, we add and subtract ${1}$, not forgetting the expression's constant term, $-39$ : $\begin{aligned} f(x)&=x^2+2x-39 \\\\ &=x^2+2x+{1}-39-{1} \\\\ &=(x+1)^2-39-1 \\\\ &=(x+1)^2-40 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 1)^2 - 40$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+2x-3 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 2x = 3$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $2$, half of it would be $1$, and squaring it gives us ${1}$. $x^2 + 2x { + 1} = 3 { + 1}$ We can now rewrite the left side of the equation as a squared term. $( x + 1 )^2 = 4$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+10x+16 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 10x = -16$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $10$, half of it would be $5$, and squaring it gives us ${25}$. $x^2 + 10x { + 25} = -16 { + 25}$ We can now rewrite the left side of the equation as a squared term. $( x + 5 )^2 = 9$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-4x+70$ $f(x)=(x+$ | We want to complete $x^2{-4}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-4}}{2}\right)^2={4}$ to it: $x^2{-4}x+{4}=(x-2)^2$ In order to keep the expression equivalent, we add and subtract ${4}$, not forgetting the expression's constant term, $70$ : $\begin{aligned} f(x)&=x^2-4x+70 \\\\ &=x^2-4x+{4}+70-{4} \\\\ &=(x-2)^2+70-4 \\\\ &=(x-2)^2+66 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 2)^2 + 66$ This is equivalent to $f(x)=(x+{-2})^2+66$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-12x-29$ $f(x)=(x+$ | We want to complete $x^2{-12}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-12}}{2}\right)^2={36}$ to it: $x^2{-12}x+{36}=(x-6)^2$ In order to keep the expression equivalent, we add and subtract ${36}$, not forgetting the expression's constant term, $-29$ : $\begin{aligned} f(x)&=x^2-12x-29 \\\\ &=x^2-12x+{36}-29-{36} \\\\ &=(x-6)^2-29-36 \\\\ &=(x-6)^2-65 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 6)^2 - 65$ This is equivalent to $f(x)=(x+{-6})^2-65$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+4x = 0$ $(x + $ | We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $4$, half of it would be $2$, and squaring it gives us ${4}$. $x^2 + 4x { + 4} = 0 { + 4}$ We can now rewrite the left side of the equation as a squared term. $( x + 2 )^2 = 4$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+8x+12 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 8x = -12$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $8$, half of it would be $4$, and squaring it gives us ${16}$. $x^2 + 8x { + 16} = -12 { + 16}$ We can now rewrite the left side of the equation as a squared term. $( x + 4 )^2 = 4$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}-12x+50$ $f(x)=(x+$ | We want to complete $x^2{-12}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-12}}{2}\right)^2={36}$ to it: $x^2{-12}x+{36}=(x-6)^2$ In order to keep the expression equivalent, we add and subtract ${36}$, not forgetting the expression's constant term, $50$ : $\begin{aligned} f(x)&=x^2-12x+50 \\\\ &=x^2-12x+{36}+50-{36} \\\\ &=(x-6)^2+50-36 \\\\ &=(x-6)^2+14 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 6)^2 + 14$ This is equivalent to $f(x)=(x+{-6})^2+14$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}+16x+63 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 + 16x = -63$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $16$, half of it would be $8$, and squaring it gives us ${64}$. $x^2 + 16x { + 64} = -63 { + 64}$ We can now rewrite the left side of the equation as a squared term. $( x + 8 )^2 = 1$ |
completing_the_square_1 | Rewrite the equation by completing the square. $x^{2}-4x+3 = 0$ $(x + $ | Begin by moving the constant term to the right side of the equation. $x^2 - 4x = -3$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-4$, half of it would be $-2$, and squaring it gives us ${4}$. $x^2 - 4x { + 4} = -3 { + 4}$ We can now rewrite the left side of the equation as a squared term. $( x - 2 )^2 = 1$ This is equivalent to $(x+{-2})^2=1$ |
completing_the_square_1 | Rewrite the function by completing the square. $f(x)=x^{2}+4x-60$ $f(x)=(x+$ | We want to complete $x^2{+4}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+4}}{2}\right)^2={4}$ to it: $x^2{+4}x+{4}=(x+2)^2$ In order to keep the expression equivalent, we add and subtract ${4}$, not forgetting the expression's constant term, $-60$ : $\begin{aligned} f(x)&=x^2+4x-60 \\\\ &=x^2+4x+{4}-60-{4} \\\\ &=(x+2)^2-60-4 \\\\ &=(x+2)^2-64 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 2)^2 - 64$ |
unit-vector | Find the unit vector in the direction of $\vec{v}=\left( 12, -5 \right)$. $($ $~,$ $)$ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 12, -5 \right) }{\sqrt{12^2+(-5)^2}} \\\\\\ &= \dfrac{1}{\sqrt{169}} \cdot \left( 12, -5 \right) \\\\\\ &= \left( {\dfrac{12}{13}}, {-\dfrac{5}{13}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{12}{13}} \right)^2 + \left( {-\dfrac{5}{13}} \right)^2} \\\\\\ &= \sqrt{\dfrac{144}{169} + \dfrac{25}{169}} \\\\\\ &= \sqrt{\dfrac{169}{169}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{12}{13}}, {-\dfrac{5}{13}}\right) $ Visualizing the answer: $12$ $-5$ $\dfrac{12}{13}$ $\dfrac{-5}{13}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Select the unit vector in the direction of $\vec{v}=\left( -7, 6 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( {-\dfrac{7}{\sqrt{85}}},{\dfrac{6}{\sqrt{85}}}\right) $ (Choice B) B $\left( {\dfrac{6}{\sqrt{85}}},{-\dfrac{7}{\sqrt{85}}}\right) $ (Choice C) C $\left( {-\dfrac{7}{\sqrt{13}}},{\dfrac{6}{\sqrt{13}}}\right) $ (Choice D) D $\left( {\dfrac{6}{\sqrt{13}}},{-\dfrac{7}{\sqrt{13}}}\right) $ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -7, 6 \right) }{\sqrt{(-7)^2+6^2}} \\\\\\ &= \dfrac{1}{\sqrt{85}} \cdot \left( -7, 6 \right) \\\\\\ &= \left( {-\dfrac{7}{\sqrt{85}}}, {\dfrac{6}{\sqrt{85}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{7}{\sqrt{85}}} \right)^2 + \left( {\dfrac{6}{\sqrt{85}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{49}{85} + \dfrac{36}{85}} \\\\\\ &= \sqrt{\dfrac{85}{85}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{7}{\sqrt{85}}}, {\dfrac{6}{\sqrt{85}}}\right) $ Visualizing the answer: $-7$ $6$ $\dfrac{-7}{\sqrt{85}}$ $\dfrac{6}{\sqrt{85}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Select the unit vector in the direction of $\vec{v}=\left( -1, -7 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( {-\dfrac{1}{\sqrt{50}}},{-\dfrac{7}{\sqrt{50}}}\right) $ (Choice B) B $\left( {\dfrac{1}{{7}}},1\right) $ (Choice C) C $\left( {-\dfrac{1}{\sqrt{-8}}},{-\dfrac{7}{\sqrt{-8}}}\right) $ (Choice D) D $\left( -1, 0\right) $ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -1, -7 \right) }{\sqrt{(-1)^2+(-7)^2}} \\\\\\ &= \dfrac{1}{\sqrt{50}} \cdot \left( -1, -7 \right) \\\\\\ &= \left( {-\dfrac{1}{\sqrt{50}}}, {-\dfrac{7}{\sqrt{50}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{1}{\sqrt{50}}} \right)^2 + \left( {-\dfrac{7}{\sqrt{50}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{1}{50} + \dfrac{49}{50}} \\\\\\ &= \sqrt{\dfrac{50}{50}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{1}{\sqrt{50}}}, {-\dfrac{7}{\sqrt{50}}}\right) $ Visualizing the answer: $-1$ $-7$ $\dfrac{-1}{\sqrt{50}}$ $\dfrac{-7}{\sqrt{50}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Select the unit vector in the direction of $\vec{v}=\left( -8, -6 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( {-\dfrac{4}{5}},{-\dfrac{3}{5}}\right) $ (Choice B) B $\left( {-\dfrac{8}{5}},{-\dfrac{6}{5}}\right) $ (Choice C) C $\left( {-\dfrac{4}{\sqrt{5}}},{-\dfrac{3}{\sqrt{5}}}\right) $ (Choice D) D $\left( {-\dfrac{8}{\sqrt{5}}},{-\dfrac{6}{\sqrt{5}}}\right) $ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -8, -6 \right) }{\sqrt{(-8)^2+(-6)^2}} \\\\\\ &= \dfrac{1}{\sqrt{100}} \cdot \left( -8, -6\right) \\\\\\ &= \left( -\dfrac{8}{10}, -\dfrac{6}{10}\right)\\\\\\ &= \left( {-\dfrac{4}{5}}, {-\dfrac{3}{5}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{4}{5}} \right)^2 + \left( {-\dfrac{3}{5}} \right)^2} \\\\\\ &= \sqrt{\dfrac{16}{25} + \dfrac{9}{25}} \\\\\\ &= \sqrt{\dfrac{25}{25}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{4}{5}}, {-\dfrac{3}{5}}\right) $ Visualizing the answer: $-8$ $-6$ $\dfrac{-8}{10}=-\dfrac{4}{5}$ $\dfrac{-6}{10}=-\dfrac{3}{5}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Select the unit vector in the direction of $\vec{v}=\left( -8, 5 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( {-\dfrac{1}{8\sqrt{89}}},{\dfrac{1}{5\sqrt{89}}}\right) $ (Choice B) B $\left( {-\dfrac{1}{5\sqrt{89}}},{\dfrac{1}{8\sqrt{89}}}\right) $ (Choice C) C $\left( {-\dfrac{8}{\sqrt{89}}},{\dfrac{5}{\sqrt{89}}}\right) $ (Choice D) D $\left( \dfrac{\sqrt{89}}{8},\dfrac{\sqrt{89}}{5}\right) $ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -8, 5 \right) }{\sqrt{(-8)^2+5^2}} \\\\\\ &= \dfrac{1}{\sqrt{89}} \cdot \left( -8, 5 \right) \\\\\\ &= \left( {-\dfrac{8}{\sqrt{89}}}, {\dfrac{5}{\sqrt{89}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{8}{\sqrt{89}}} \right)^2 + \left( {\dfrac{5}{\sqrt{89}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{64}{89} + \dfrac{25}{89}} \\\\\\ &= \sqrt{\dfrac{89}{89}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{8}{\sqrt{89}}}, {\dfrac{5}{\sqrt{89}}}\right) $ Visualizing the answer: $-8$ $5$ $\dfrac{-8}{\sqrt{89}}$ $\dfrac{5}{\sqrt{89}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Select the unit vector in the direction of $\vec{v}=\left( -3, 8 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( {-\dfrac{3}{\sqrt{55}}},{\dfrac{8}{\sqrt{55}}}\right) $ (Choice B) B $\left( {-\dfrac{3}{\sqrt{73}}},{\dfrac{8}{\sqrt{73}}}\right) $ (Choice C) C $\left( {-\dfrac{3}{\sqrt{11}}},{\dfrac{8}{\sqrt{11}}}\right) $ (Choice D) D $\left( {-\dfrac{3}{\sqrt{5}}},{\dfrac{8}{\sqrt{5}}}\right) $ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -3, 8 \right) }{\sqrt{(-3)^2+8^2}} \\\\\\ &= \dfrac{1}{\sqrt{73}} \cdot \left( -3, 8 \right) \\\\\\ &= \left( {-\dfrac{3}{\sqrt{73}}}, {\dfrac{8}{\sqrt{73}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{3}{\sqrt{73}}} \right)^2 + \left( {\dfrac{8}{\sqrt{73}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{9}{73} + \dfrac{64}{73}} \\\\\\ &= \sqrt{\dfrac{73}{73}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{3}{\sqrt{73}}}, {\dfrac{8}{\sqrt{73}}}\right) $ Visualizing the answer: $-3$ $8$ $\dfrac{-3}{\sqrt{73}}$ $\dfrac{8}{\sqrt{73}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Find the unit vector in the direction of $\vec{v}=\left( 1, -2 \right)$. $($ $~,$ $)$ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 1, -2 \right) }{\sqrt{1^2+(-2)^2}} \\\\\\ &= \dfrac{1}{\sqrt{5}} \cdot \left( 1, -2 \right) \\\\\\ &= \left( {\dfrac{1}{\sqrt{5}}}, {-\dfrac{2}{\sqrt{5}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{1}{\sqrt{5}}} \right)^2 + \left( {-\dfrac{2}{\sqrt{5}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{1}{5} + \dfrac{4}{5}} \\\\\\ &= \sqrt{\dfrac{5}{5}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{1}{\sqrt{5}}}, {-\dfrac{2}{\sqrt{5}}}\right) $ Visualizing the answer: $1$ $-2$ $\dfrac{1}{\sqrt{5}}$ $\dfrac{-2}{\sqrt{5}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Find the unit vector in the direction of $\vec{v}=\left( 3,4 \right)$. $($ $~,$ $)$ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 3, 4 \right) }{\sqrt{3^2+4^2}} \\\\\\ &= \dfrac{1}{\sqrt{25}} \cdot \left( 3, 4 \right) \\\\\\ &= \left( {\dfrac{3}{5}}, {\dfrac{4}{5}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{3}{5}} \right)^2 + \left( {\dfrac{4}{5}} \right)^2} \\\\\\ &= \sqrt{\dfrac{9}{25} + \dfrac{16}{25}} \\\\\\ &= \sqrt{\dfrac{25}{25}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{3}{5}}, {\dfrac{4}{5}}\right) $ Visualizing the answer: $3$ $4$ $\dfrac{3}{5}$ $\dfrac{4}{5}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Find the unit vector in the direction of $\vec{v}=\left( 6, 3 \right)$. $($ $~,$ $)$ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 6, 3 \right) }{\sqrt{6^2+3^2}} \\\\\\ &= \dfrac{1}{\sqrt{45}} \cdot \left( 6, 3 \right) \\\\\\ &= \left( {\dfrac{6}{\sqrt{45}}}, {\dfrac{3}{\sqrt{45}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{6}{\sqrt{45}}} \right)^2 + \left( {\dfrac{3}{\sqrt{45}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{36}{45} + \dfrac{9}{45}} \\\\\\ &= \sqrt{\dfrac{45}{45}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{6}{\sqrt{45}}}, {\dfrac{3}{\sqrt{45}}}\right) $ Visualizing the answer: $6$ $3$ $\dfrac{6}{\sqrt{45}}$ $\dfrac{3}{\sqrt{45}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Find the unit vector in the direction of $\vec{v}=\left( 5, -2\right)$. $($ $~,$ $)$ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 5, -2 \right) }{\sqrt{5^2+(-2)^2}} \\\\\\ &= \dfrac{1}{\sqrt{29}} \cdot \left( 5, -2 \right) \\\\\\ &= \left( {\dfrac{5}{\sqrt{29}}}, {-\dfrac{2}{\sqrt{29}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{5}{\sqrt{29}}} \right)^2 + \left( {-\dfrac{2}{\sqrt{29}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{25}{29} + \dfrac{4}{29}} \\\\\\ &= \sqrt{\dfrac{29}{29}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{5}{\sqrt{29}}}, {-\dfrac{2}{\sqrt{29}}}\right) $ Visualizing the answer: $5$ $-2$ $\dfrac{5}{\sqrt{29}}$ $\dfrac{-2}{\sqrt{29}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Find the unit vector in the direction of $\vec{v}=\left( 8, 9 \right)$. $($ $~,$ $)$ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 8, 9 \right) }{\sqrt{8^2+9^2}} \\\\\\ &= \dfrac{1}{\sqrt{145}} \cdot \left( 8, 9 \right) \\\\\\ &= \left( {\dfrac{8}{\sqrt{145}}}, {\dfrac{9}{\sqrt{145}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{8}{\sqrt{145}}} \right)^2 + \left( {\dfrac{9}{\sqrt{145}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{64}{145} + \dfrac{81}{145}} \\\\\\ &= \sqrt{\dfrac{145}{145}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{8}{\sqrt{145}}}, {\dfrac{9}{\sqrt{145}}}\right) $ Visualizing the answer: $8$ $9$ $\dfrac{8}{\sqrt{145}}$ $\dfrac{9}{\sqrt{145}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Select the unit vector in the direction of $\vec{v}=\left( -2, 3 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( -2, 3\right) $ (Choice B) B $\left( {\dfrac{2}{\sqrt{6}}},{-\dfrac{3}{\sqrt{6}}}\right) $ (Choice C) C $\left( 0, 3 \right) $ (Choice D) D $\left( {-\dfrac{2}{\sqrt{13}}},{\dfrac{3}{\sqrt{13}}}\right) $ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -2, 3 \right) }{\sqrt{(-2)^2+3^2}} \\\\\\ &= \dfrac{1}{\sqrt{13}} \cdot \left( -2, 3 \right) \\\\\\ &= \left( {-\dfrac{2}{\sqrt{13}}}, {\dfrac{3}{\sqrt{13}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{2}{\sqrt{13}}} \right)^2 + \left( {\dfrac{3}{\sqrt{13}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{4}{13} + \dfrac{9}{13}} \\\\\\ &= \sqrt{\dfrac{13}{13}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{2}{\sqrt{13}}}, {\dfrac{3}{\sqrt{13}}}\right) $ Visualizing the answer: $-2$ $3$ $\dfrac{-2}{\sqrt{13}}$ $\dfrac{3}{\sqrt{13}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Find the unit vector in the direction of $\vec{v}=\left( 2, -8 \right)$. $($ $~,$ $)$ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 2, -8 \right) }{\sqrt{2^2+(-8)^2}} \\\\\\ &= \dfrac{1}{\sqrt{68}} \cdot \left( 2, -8 \right) \\\\\\ &= \left( {\dfrac{2}{\sqrt{68}}}, {-\dfrac{8}{\sqrt{68}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{2}{\sqrt{68}}} \right)^2 + \left( {-\dfrac{8}{\sqrt{68}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{4}{68} + \dfrac{64}{68}} \\\\\\ &= \sqrt{\dfrac{68}{68}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{2}{\sqrt{68}}}, {-\dfrac{8}{\sqrt{68}}}\right) $ Visualizing the answer: $2$ $-8$ $\dfrac{2}{\sqrt{68}}$ $\dfrac{-8}{\sqrt{68}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Select the unit vector in the direction of $\vec{v}=\left( -5, 5 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( -1, 1\right) $ (Choice B) B $\left( {-\dfrac{5}{\sqrt{50}}},{\dfrac{5}{\sqrt{50}}}\right) $ (Choice C) C $\left( -5, 0 \right) $ (Choice D) D $\left( 0, 5 \right) $ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -5, 5 \right) }{\sqrt{(-5)^2+5^2}} \\\\\\ &= \dfrac{1}{\sqrt{50}} \cdot \left( -5, 5 \right) \\\\\\ &= \left( {-\dfrac{5}{\sqrt{50}}}, {\dfrac{5}{\sqrt{50}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{5}{\sqrt{50}}} \right)^2 + \left( {\dfrac{5}{\sqrt{50}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{25}{50} + \dfrac{25}{50}} \\\\\\ &= \sqrt{\dfrac{50}{50}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{5}{\sqrt{50}}}, {\dfrac{5}{\sqrt{50}}}\right) $ Visualizing the answer: $-5$ $5$ $\dfrac{-5}{\sqrt{50}}$ $\dfrac{5}{\sqrt{50}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Select the unit vector in the direction of $\vec{v}=\left( -2, -2 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( {-\dfrac{2}{\sqrt{-4}}},{-\dfrac{2}{\sqrt{-4}}}\right) $ (Choice B) B $\left( {\dfrac{1}{{2}}},{\dfrac{1}{{2}}}\right) $ (Choice C) C $\left( {-\dfrac{2}{\sqrt{8}}},{-\dfrac{2}{\sqrt{8}}}\right) $ (Choice D) D $\left( {-\dfrac{2}{{8}}},{-\dfrac{2}{{8}}}\right) $ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -2, -2 \right) }{\sqrt{(-2)^2+(-2)^2}} \\\\\\ &= \dfrac{1}{\sqrt{8}} \cdot \left( -2, -2 \right) \\\\\\ &= \left( {-\dfrac{2}{\sqrt{8}}}, {-\dfrac{2}{\sqrt{8}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{2}{\sqrt{8}}} \right)^2 + \left( {-\dfrac{2}{\sqrt{8}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{4}{8} + \dfrac{4}{8}} \\\\\\ &= \sqrt{\dfrac{8}{8}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{2}{\sqrt{8}}}, {-\dfrac{2}{\sqrt{8}}}\right) $ Visualizing the answer: $-2$ $-2$ $\dfrac{-2}{\sqrt{8}}$ $\dfrac{-2}{\sqrt{8}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Find the unit vector in the direction of $\vec{v}=\left( 6, 6 \right)$. $($ $~,$ $)$ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 6, 6 \right) }{\sqrt{6^2+6^2}} \\\\\\ &= \dfrac{1}{\sqrt{72}} \cdot \left( 6, 6 \right) \\\\\\ &= \left( {\dfrac{6}{\sqrt{72}}}, {\dfrac{6}{\sqrt{72}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{6}{\sqrt{72}}} \right)^2 + \left( {\dfrac{6}{\sqrt{72}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{36}{72} + \dfrac{36}{72}} \\\\\\ &= \sqrt{\dfrac{72}{72}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{6}{\sqrt{72}}}, {\dfrac{6}{\sqrt{72}}}\right) $ Visualizing the answer: $6$ $6$ $\dfrac{6}{\sqrt{72}}$ $\dfrac{6}{\sqrt{72}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Select the unit vector in the direction of $\vec{v}=\left( -4, -8 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( {\dfrac{4}{\sqrt{80}}},{-\dfrac{8}{\sqrt{80}}}\right) $ (Choice B) B $\left( {-\dfrac{4}{\sqrt{80}}},{\dfrac{8}{\sqrt{80}}}\right) $ (Choice C) C $\left( {-\dfrac{4}{\sqrt{80}}},{-\dfrac{8}{\sqrt{80}}}\right) $ (Choice D) D $\left( {\dfrac{4}{\sqrt{80}}},{\dfrac{8}{\sqrt{80}}}\right) $ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -4, -8 \right) }{\sqrt{(-4)^2+(-8)^2}} \\\\\\ &= \dfrac{1}{\sqrt{80}} \cdot \left( -4, -8 \right) \\\\\\ &= \left( {-\dfrac{4}{\sqrt{80}}}, {-\dfrac{8}{\sqrt{80}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{4}{\sqrt{80}}} \right)^2 + \left( {-\dfrac{8}{\sqrt{80}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{16}{80} + \dfrac{64}{80}} \\\\\\ &= \sqrt{\dfrac{80}{80}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{4}{\sqrt{80}}}, {-\dfrac{8}{\sqrt{80}}}\right) $ Visualizing the answer: $-4$ $-8$ $\dfrac{-4}{\sqrt{80}}$ $\dfrac{-8}{\sqrt{80}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
unit-vector | Find the unit vector in the direction of $\vec{v}=\left( 7, -3 \right)$. Enter exact answers with square roots. $($ $~,$ $)$ | Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 7, -3 \right) }{\sqrt{7^2+(-3)^2}} \\\\\\ &= \dfrac{1}{\sqrt{58}} \cdot \left( 7, -3 \right) \\\\\\ &= \left( {\dfrac{7}{\sqrt{58}}}, {-\dfrac{3}{\sqrt{58}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{7}{\sqrt{58}}} \right)^2 + \left( {-\dfrac{3}{\sqrt{58}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{49}{58} + \dfrac{9}{58}} \\\\\\ &= \sqrt{\dfrac{58}{58}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{7}{\sqrt{58}}}, {-\dfrac{3}{\sqrt{58}}}\right) $ Visualizing the answer: $7$ $-3$ $\dfrac{7}{\sqrt{58}}$ $\dfrac{-3}{\sqrt{58}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$. |
Subsets and Splits