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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Determine The Equation Of A Parabola And Graph A Parabola

Express the following equation for a parabola in standard form:

$\frac{x^2-6x}{3}=y+7$

Possible Answers:

$y=\frac{7(x^2-6x)}{3}$

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

$y=-\frac{7}{3}x^2-2x-\frac{7}{3}$

$y=\frac{7}{3}x^2+2x+\frac{7}{3}$

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

Correct answer:

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

Explanation:

In order to be in standard form, the equation for a parabola must be written in one of the following ways:

y=ax^2+bx+c

x=ay^2+by+c

We can see that our equation has all of the components of the first form above, so now all we must do is some algebra to rearrange the equation and express the function y in terms of x. We start by simplifying the fraction on the left side of the equation, and then we isolate y to give us the equation of the parabola in standard form:

$\frac{x^2-6x}{3}=y+7$

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

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

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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+15x+11

y= 25x^2+11

y= 25x^2+30x+9

y= 25x^2+30x+11

y= 25x^2+15x+2

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 #231 : Conic Sections

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

Possible Answers:

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

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

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

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

y=-x^2-2x+4

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 #232 : Conic Sections

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 #1811 : Pre Calculus

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 #231 : Conic Sections

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)

Report an Error

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.

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

Report an Error

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)

Report an Error

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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #3 : Determine The Equation Of A Parabola And Graph A Parabola

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

Example Question #231 : Conic Sections

Example Question #232 : Conic Sections

Example Question #1811 : Pre Calculus

Example Question #231 : Conic Sections

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

All Precalculus Resources

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