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Precalculus : Parabolas

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Example Questions

Example Question #41 : Parabolas

Rewrite the following equation for a parabola in standard form:

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

Possible Answers:

y=13-4x^2

y=4x^2-16x+16

y=4x^2-16x+19

y=x^2-4x+7

y=4x^2-13

Correct answer:

y=4x^2-16x+19

Explanation:

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

y=ax^2+bx+c OR x=ay^2+by+c

THe problem given has the square around the x term, so it's going to end up loking like the standard form on the left.

First, we square the right side

y-3=(2x-4)(2x-4)

y-3=4x^2-8x-8x+16

y-3=4x^2-16x+16

y-3+3=4x^2-16x+16+3

Lastly, we need the y by itself, so we add 3 to both sides

y-3+3=4x^2-16x+16+3

y=4x^2-16x+19

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Example Question #42 : Parabolas

Write the equation for a vertex of (5,7) that passes through the origin.

Possible Answers:

$y = -\frac{3}{14}(x-5)^2-7$

$y = -\frac{2}{10}(x+5)^2+7$

$y = -\frac{7}{25}(x-5)^2+7$

$y = -\frac{8}{25}(x-5)^2+7$

$y = -\frac{12}{25}(x-5)^2+7$

Correct answer:

$y = -\frac{7}{25}(x-5)^2+7$

Explanation:

The vertex form for a parabola is given below:

y = a(x-h)^2+k

The vertex coordinates are (h,k) or (5,7) as given in the problem statement.

y = a(x-5)^2+7

Now solve for a using the point it crosses (0,0) , by plugging the point into the equation.

(0) = a((0)-5)^2+7

-7 = 25a

$a = -\frac{7}{25}$

Plug a back to find the final answer:

$y = -\frac{7}{25}(x-5)^2+7$

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Example Question #43 : Parabolas

Write the equation for a vertex of (4,1) that passes through the point (1,2) .

Possible Answers:

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

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

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

$y = -\frac{5}{16}(x-4)^2+1$

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

Correct answer:

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

Explanation:

The vertex form for a parabola is given below:

y = a(x-h)^2+k

The vertex coordinates are (h,k) or (4,1) as given in the problem statement.

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

Now solve for a using the point it crosses (1,2) , by plugging the point into the equation.

(2) = a((1)-4)^2+7

-5 = 9a

$a = -\frac{5}{9}$

Plug a back to find the final answer:

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

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Example Question #44 : Parabolas

Write the equation for a vertex of (-4,2) that passes through the point (4,7) .

Possible Answers:

$y = \frac{5}{72}(x+4)^2+2$

$y = \frac{5}{7}(x+4)^2+2$

$y = \frac{5}{64}(x+4)^2+2$

$y = \frac{5}{108}(x+4)^2+2$

$y = \frac{5}{8}(x+4)^2+2$

Correct answer:

$y = \frac{5}{64}(x+4)^2+2$

Explanation:

The vertex form for a parabola is given below:

y = a(x-h)^2+k

The vertex coordinates are (h,k) or (-4,2) as given in the problem statement.

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

Now solve for a using the point it crosses (4,7) , by plugging the point into the equation.

(7) = a((4)+4)^2+2

5 = 64a

$a = \frac{5}{64}$

Plug a back to find the final answer:

$y = \frac{5}{64}(x+4)^2+2$

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Example Question #1831 : Pre Calculus

Find the focus and the directrix of the following parabola: y^2=16x .

Possible Answers:

Focus: (-4,0)

Directrix: x=4

Focus: (4, 0)
Directrix: x=-4

Focus: (0,-4)

Directrix: x=-4

Focus: (0,4)

Directrix: x=4

Correct answer:

Focus: (4, 0)
Directrix: x=-4

Explanation:

To find the focus from the equation of a parabola, first set the equation to resemble the form y^2=4px where represents any numerical value.

For our problem, it is already in this form.

Therefore,

16=4p .

Solve for p: 16=4p then 4=p .

The focus for this parabola is given by (0,p) .

So, (0,4) is the focus of the parabola.

The directrix is represented as x= -p .

Therefore, the directrix for this problem is x=-4 .

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Example Question #1832 : Pre Calculus

Find the directerix of the parabola with the following equation:

(x-5)^2=20y

Possible Answers:

x=-5

y=5

y=-5

x=5

Correct answer:

y=-5

Explanation:

Recall the standard form of the equation of a vertical parabola:

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

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

For the parabola in question, the vertex is (5, 0) and p=5 . This parabola will open up. Because the parabola will open up, the directerix will be located units down from the vertex. The equation for the directerix is then y=-5 .

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Example Question #251 : Conic Sections

Find the directerix for the parabola with the following equation:

(x-5)^2=4(y+1)

Possible Answers:

x=2

y=2

y=-2

x=-2

Correct answer:

y=-2

Explanation:

Recall the standard form of the equation of a vertical parabola:

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

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

For the parabola in question, the vertex is (5, -1) and p=1 . This parabola will open up. Because the parabola will open up, the directerix will be located unit down from the vertex. The equation for the directerix is then y=-2 .

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Example Question #252 : Conic Sections

Find the directerix of the parabola with the following equation:

x^2-4x+16y-76=0

Possible Answers:

y=9

x=1

x=9

y=1

Correct answer:

y=9

Explanation:

Recall the standard form of the equation of a vertical parabola:

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

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

Start by putting the equation in th estandard form of the equation of a vertical parabola.

Isolate the terms to one side.

x^2-4x+16y-76=0

x^2-4x=-16y+76

Complete the square for the terms. Remember to add the same amount on both sides!

x^2-4x+4=-16y+76+4

x^2-4x+4=-16y+80

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

(x-2)^2=-16(y-5)

For the parabola in question, the vertex is (2, 5) and p=-4 . This parabola will open down. Because the parabola will open down, the directerix will be located units above the vertex. The equation for the directerix is then y=9 .

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Example Question #35 : Determine The Equation Of A Parabola And Graph A Parabola

Find the focus of the parabola with the following equation:

x^2+10x-4y+17=0

Possible Answers:

(5, 4)

(-5, -1)

(-5, 1)

(5, 2)

Correct answer:

(-5, -1)

Explanation:

Recall the standard form of the equation of a vertical parabola:

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

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

Start by putting the equation into the standard form.

Isolate the terms on one side.

x^2+10x-4y+17=0

x^2+10x=4y-17

Complete the square. Remember to add teh same amount on both sides!

x^2+10x+25=4y-17+25

x^2+10x+25=4y+8

Factor both sides of the equation to get the standard equation for the parabola.

(x+5)^2=4(y+2)

For the parabola in question, the vertex is (-5, -2) and p=1 . This parabola will open up. Because the parabola will open up, the focus will be located unit up from the vertex. The focus is then located at (-5, -1) .

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Example Question #33 : Determine The Equation Of A Parabola And Graph A Parabola

Find the focus of the parabola with the following equation:

(x-9)^2=-8(y-5)

Possible Answers:

(9, 11)

(9, 5)

(9, 3)

(9, -3)

Correct answer:

(9, 3)

Explanation:

Recall the standard form of the equation of a vertical parabola:

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

When p>0 , the parabola will open up.

When p<0 , the parabola will open down.

For the parabola in question, the vertex is (9, 5) and p=-2 . This parabola will open down. Because the parabola will open down, the focus will be located units down from the vertex. The focus is then located at (9, 3) .

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Precalculus : Parabolas

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #41 : Parabolas

Example Question #42 : Parabolas

Example Question #43 : Parabolas

Example Question #44 : Parabolas

Example Question #1831 : Pre Calculus

Example Question #1832 : Pre Calculus

Example Question #251 : Conic Sections

Example Question #252 : Conic Sections

Example Question #35 : Determine The Equation Of A Parabola And Graph A Parabola

Example Question #33 : Determine The Equation Of A Parabola And Graph A Parabola

All Precalculus Resources

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