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Precalculus : Determine the equation of a parabola and graph a parabola

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Example Questions

Example Question #11 : Determine The Equation Of A Parabola And Graph A Parabola

Rewrite y-2=(5x+3)^2 in standard form.

Possible Answers:

y= 25x^2+30x+11

y= 25x^2+11

y= 25x^2+15x+11

y= 25x^2+15x+2

y= 25x^2+30x+9

Correct answer:

y= 25x^2+30x+11

Explanation:

The standard form of a parabola is y=ax^2+bx+c .

Factorize the right side of y-2=(5x+3)^2 , and simplify.

(5x+3)^2=(5x+3)(5x+3)= 25x^2+30x+9

y-2= 25x^2+30x+9

y= 25x^2+30x+11

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Example Question #12 : Determine The Equation Of A Parabola And Graph A Parabola

If the vertex of the parabola is (-1,5 ) , and the y-intercept is , find the equation of the parabola, if possible.

Possible Answers:

$y=\frac{1}{5}x^2+x+4$

$y=\frac{1}{3}x^2+4$

$y=-\frac{1}{3}x^2-\frac{2}{3}x+4$

y=-x^2-2x+4

$\textup{There is not enough information to solve for the equation.}$

Correct answer:

y=-x^2-2x+4

Explanation:

First, write the equation of the parabola in standard form.

y=ax^2+bx+c

Determine the values of the coefficients. The value of the y-intercept is 4, which means that c=4 .

Write the vertex formula.

$x=-\frac{b}{2a}$

The given point of the vertex is (-1,5 ) , which indicates that:

$-\frac{b}{2a}=-1$

b=2a

Substitute the value of the point and b=2a into the standard form.

5=a(-1)^2+(2a)(-1)+4

5=a-2a+4

1=-a

a=-1

Substitute this value into b=2a to determine the value of .

b=2a=2(-1)= -2

Substitute the values of a,b,c coefficients into the standard form of the parabola.

y=-x^2-2x+4

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Example Question #13 : Determine The Equation Of A Parabola And Graph A Parabola

Find the standard form of the equation of the following parabola:

y^2-4y-4x+8=0

Possible Answers:

(y-4)^2=4(x-1)

(y-2)^2=4(x-1)

(y-1)^2=x+4

(y+2)^2=4(x+1)

Correct answer:

(y-2)^2=4(x-1)

Explanation:

Recall the standard equation of a horizontal parabola:

(y-k)^2=4p(x-h) , where (h,k) is the vertex and |p| is the focal length.

Start by isolating the terms.

y^2-4y-4x+8=0

y^2-4y=4x-8

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

y^2-4y+4=4x-8+4

y^2-4y+4=4x-4

Factor both sides of the equation to get the standard form of a horizontal parabola.

(y-2)^2=4(x-1)

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Example Question #14 : Determine The Equation Of A Parabola And Graph A Parabola

Find the standard form of equation of the following parabola:

y^2-2y+36x-35=0

Possible Answers:

(y+1)^2=36(x+1)

(y-1)^2=-36(x-1)

(y-1)^2=36(x-1)

(y+1)^2=-36(x+1)

Correct answer:

(y-1)^2=-36(x-1)

Explanation:

Recall the standard equation of a horizontal parabola:

(y-k)^2=4p(x-h) , where (h,k) is the vertex and |p| is the focal length.

Start by isolating the terms.

y^2-2y+36x-35=0

y^2-2y=-36x+35

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

y^2-2y+1=-36x+35+1

y^2-2y+1=-36x+36

Factor both sides of the equation to get the standard form of a horizontal parabola.

(y-1)^2=-36(x-1)

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Example Question #15 : Determine The Equation Of A Parabola And Graph A Parabola

Find the standard form of the equation for the following parabola:

y^2+8y+6x+10=0

Possible Answers:

(y+4)^2=6(x-1)

(y+4)^2=-6(x-1)

(y-4)^2=-6(x-1)

(y-1)^2=-6(x+4)

Correct answer:

(y+4)^2=-6(x-1)

Explanation:

Recall the standard equation of a horizontal parabola:

(y-k)^2=4p(x-h) , where (h,k) is the vertex and |p| is the focal length.

Start by isolating the terms.

y^2+8y+6x+10=0

y^2+8y=-6x-10

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

y^2+8y+16=-6x-10+16

y^2+8y+16=-6x+6

Factor both sides of the equation to get the standard form of a horizontal parabola.

(y+4)^2=-6(x-1)

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Example Question #16 : Determine The Equation Of A Parabola And Graph A Parabola

Find the standard form of the equation for the following parabola:

$x^2-14x-\frac{1}{2}y+47=0$

Possible Answers:

$(x-7)^2=\frac{1}{2}(y-8)$

$(x+7)^2=\frac{1}{2}(y-3)$

$(x-7)^2=\frac{1}{2}(y+4)$

$(x+4)^2=\frac{1}{2}(x-2)$

Correct answer:

$(x-7)^2=\frac{1}{2}(y+4)$

Explanation:

Recall the standard equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h,k) is the vertex and |p| is the focal length.

Start by isolating the terms.

$x^2-14x-\frac{1}{2}y+47=0$

$x^2-14x=\frac{1}{2}y-47$

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

$x^2-14x+49=\frac{1}{2}y-47+49$

$x^2-14x+49=\frac{1}{2}y+2$

Factor both sides of the equation to get the standard form of a vertical parabola.

$(x-7)^2=\frac{1}{2}(y+4)$

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Example Question #16 : Determine The Equation Of A Parabola And Graph A Parabola

Find the standard form of the equation for the following parabola:

$x^2+6x-\frac{1}{4}y+5=0$

Possible Answers:

(x+3)^2=4(y+4)

(x+3)^2=4(y-1)

$(x-3)^2=\frac{1}{4}(y-8)$

$(x+3)^2=\frac{1}{4}(y+16)$

Correct answer:

$(x+3)^2=\frac{1}{4}(y+16)$

Explanation:

Recall the standard equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h,k) is the vertex and |p| is the focal length.

Start by isolating the terms.

$x^2+6x-\frac{1}{4}y+5=0$

$x^2+6x=\frac{1}{4}y-5$

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

$x^2+6x+9=\frac{1}{4}y-5+9$

$x^2+6x+9=\frac{1}{4}y+4$

Factor both sides of the equation to get the standard form of a vertical parabola.

$(x+3)^2=\frac{1}{4}(y+16)$

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Example Question #31 : Parabolas

Find the standard form of the equation for the following parabola:

x^2-20x-24y+28=0

Possible Answers:

(x+10)^2=24(y+2)

(x+10)^2=24(y-1)

(x-10)^2=24(y-3)

(x-10)^2=24(y+3)

Correct answer:

(x-10)^2=24(y+3)

Explanation:

Recall the standard equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h,k) is the vertex and |p| is the focal length.

Start by isolating the terms.

x^2-20x-24y+28=0

x^2-20x=24y-28

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

x^2-20x+100=24y-28+100

x^2-20x+100=24y+72

Factor both sides of the equation to get the standard form of a vertical parabola.

(x-10)^2=24(y+3)

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Example Question #18 : Determine The Equation Of A Parabola And Graph A Parabola

Find the standard form of the equation of a parabola with the following equation:

x^2-18x+100y-119=0

Possible Answers:

(x-9)^2=-100(y-2)

(x-9)^2=-100(y+2)

(x+2)^2=-100(y-9)

(x-2)^2=100(y-9)

Correct answer:

(x-9)^2=-100(y-2)

Explanation:

Recall the standard equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h,k) is the vertex and |p| is the focal length.

Start by isolating the terms.

x^2-18x+100y-119=0

x^2-18x=-100y+119

Complete the square on the left. Make sure to add the same amount to both sides of the equation!

x^2-18x+81=-100y+119+81

x^2-18x+81=-100y+200

Factor both sides of the equation to get the standard form of a vertical parabola.

(x-9)^2=-100(y-2)

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Example Question #14 : Determine The Equation Of A Parabola And Graph A Parabola

Find the standard form of the equation of a parabola that has its focus at (1, 4) and its directerix at y=-2 .

Possible Answers:

(x-1)^2=12(x-2)

(x-1)^2=3(y-1)

(x-1)^2=12(y-1)

(x+1)^2=12(y+1)

Correct answer:

(x-1)^2=12(y-1)

Explanation:

First, we need to figure out if this is a vertical parabola or a horizontal parabola. Since we need to move down units from the focus to the directerix, we know that this is a vertical parabola.

Recall the standard equation of a vertical parabola:

(x-h)^2=4p(y-k) , where (h,k) is the vertex and |p| is the focal length.

Remember that the distance from the focus to the directerix is 2|p| . Then, |p|=3 .

With this information, we can then figure out the vertex of the parabola. Since this is a vertical parabola, the -coordinate of the focus and the vertex are the same. To find the -coordinate of the vertex, just move down units from the focus. The vertex is then (1, 1) .

Since the focus is above the vertex, we know that p=3 .

Now, from the vertex, we know that h=1 and k=1 .

Now, plug in all the information to write the standard equation for this parabola:

(x-1)^2=12(y-1)

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Precalculus : Determine the equation of a parabola and graph a parabola

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #11 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #12 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #13 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #14 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #15 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #16 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #16 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #31 : Parabolas

Example Question #18 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #14 : Determine The Equation Of A Parabola And Graph A Parabola

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