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College Algebra : Parabolas

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Example Questions

Example Question #1 : Parabolas

Which of the following would give a graph of an upward facing parabola?

Possible Answers:

y=-4x^2+13x+156

y=3x^2-4x-12

y=3x^3-4x+2

$y=e^{3x}+12x$

Correct answer:

y=3x^2-4x-12

Explanation:

Which of the following would give a graph of an upward facing parabola?

Parabolas are created from polynomials where the highest exponent is 2.

To be an upward facing parabola, the squared term must be positive.

The only option that matches these criteria is:

y=3x^2-4x-12

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Example Question #2 : Parabolas

Find the coordinates of the vertex of this quadratic equation: 6x^2+24x+18

Possible Answers:

(-2,-6)

This is not a parabola.

(-2,6)

(2,6)

(6,2)

Correct answer:

(-2,-6)

Explanation:

To find the vertex of this parabola use the following formula to find the x-coordinate of its vertex; find the y-coordinate by substituting the x-coordinate into the equation.

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

6x^2+24x+18

$\frac{-24}{2(6)}=\boldsymbol{-2}$

To find the y-coordinate, substitute -2 back into the quadratic equation:

$6x^2+24x+18\rightarrow 6(-2)^2+24(-2)+18=\boldsymbol{-6}$

The vertex is $\boldsymbol{(-2,-6)}$ .

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Example Question #3 : Parabolas

How many -intercepts does the graph of the function

f(x)= 3x^2+12x+8

have?

Possible Answers:

Two

Zero

One

Correct answer:

Two

Explanation:

The graph of a quadratic function $f(x) = ax^{2}+ bx + c$ has an -intercept at any point (x,0 ) at which f(x) = 0 , so, first, set the quadratic expression equal to 0:

3x^2+12x+8 =0

The number of -intercepts of the graph is equal to the number of real zeroes of the above equation, which can be determined by evaluating the discriminant of the equation, $b^{2} -4ac$ . Set a =3, b = 12, c = 8 , and evaluate:

$b^{2} -4ac = 12^{2} - 4 (3)(8) = 14 4 - 96 = 48$

The discriminant is positive, so the equation has two solutions, both of which are real. Consequently, the graph of the function f(x) has two -intercepts

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Example Question #2 : Parabolas

A baseball is thrown off the roof of a building 320 feet high at an initial upward speed of 80 feet per second; the height of the baseball relative to the ground is modeled by the function

$h(t) = -16t^{2}+ 80t + 320$

How long does it take for the baseball to reach its highest point (nearest tenth of a second)?

Possible Answers:

$5.2\textrm{ sec}$

$7.6\textrm{ sec}$

$4.0\textrm{ sec}$

$2.5 \textrm{ sec}$

$4.2\textrm{ sec}$

Correct answer:

$2.5 \textrm{ sec}$

Explanation:

The highest point of the ball is the vertex of the ball's parabolic path, so to find the number of seconds that is takes to reach this point, it is necessary to find the first coordinate of the vertex of the parabola of the graph of the function

$h(t) = -16t^{2}+ 80t + 320$

The parabola of the graph of

$h(t) =at^{2}+ bt+ c$

has as its ordinate, or -coordinate,

$t = -\frac{b}{2a}$ ,

so, setting a = -16, b = 80 ,

$t = -\frac{80}{2 (-16)} = -\frac{80}{-32} = 2.5$ ,

This is the time in seconds that it takes the ball to reach the highest point of its path.

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Example Question #2 : Parabolas

Give the vertex of the parabola of the equation

$x= y^{2}+ 8y - 12$

Possible Answers:

(36, -12)

(36, 4)

(-12, 0)

(228, 12)

$\left ( -28 , -4\right )$

Correct answer:

$\left ( -28 , -4\right )$

Explanation:

The parabola of the equation $x = ay^{2}+ by+ c$ has its vertex at a point with -coordinate $- \frac{b}{2a}$ ; set a = 1 and b = 8 in this formula to get

$- \frac{8}{2(1)} = -4$

Substitute this for in the equation to obtain the corresponding -coordinate:

$x= y^{2}+ 8y - 12$

$x= ( -4)^{2}+ 8( -4) - 12 = 16 + (-32 )- 12 = -28$

The vertex is at $\left ( -28 , -4\right )$ .

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Example Question #2 : Parabolas

What is the multiplicity of the root of a quadratic equation with a discriminant equal to 0?

Possible Answers:

Correct answer:

Explanation:

Quadratic equations whose discriminants are equal to zero have one repeated root (solution). Because this root appears twice in the quadratic equation, it has a multiplicity of 2. The number of times a factor appears in a polynomial, such as quadratic, is its multiplicity.

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Example Question #3 : Parabolas

Define a function $f(x) = -4x^{2}+ 8x -2$ .

Which of the following is the -coordinate of the -intercept of its graph?

Possible Answers:

Correct answer:

Explanation:

The -intercept of the graph of a function f(x) is the point at which it crosses the -axis; its -coordinate is 0, so its -coordinate is f(0) .

$f(x) = -4x^{2}+ 8x -2$ ,

so, by setting x= 0 ,

$f(0) = -4 \cdot 0^{2}+ 8 \cdot 0 -2 = 0+0-2 = -2$

The -intercept is (0,-2) .

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Example Question #25 : Graphs

Try without a calculator.

The graph of the function

f(x)= - 3x^2+12x+15

is a parabola. Which choice correctly gives its concavity?

Possible Answers:

Concave to the right

Concave to the left

Concave downward

Concave upward

Correct answer:

Concave downward

Explanation:

The direction of the concavity of the parabola of the function

$f(x) = ax^{2}+ bx + c$

is either upward or downward depending entirely on the sign of , the coefficient of $x^{2}$ . This coefficient, , is negative; the parabola is concave downward.

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Example Question #5 : Parabolas

Give the coordinates of the focus of the parabola of the equation

$y= 4x^{2} + 4x+ 8$

Possible Answers:

$\left ( -\frac{1}{2}, \frac{112}{16} \right )$

$\left ( -\frac{1}{2}, \frac{113}{16} \right )$

$\left ( -\frac{9}{16} , 7 \right )$

$\left ( -\frac{1}{2} , 7 \right )$

$\left ( -\frac{7}{16} , 7 \right )$

Correct answer:

$\left ( -\frac{1}{2}, \frac{113}{16} \right )$

Explanation:

The parabola in question is a vertical parabola. Its equation is in the standard form

$y = ax^{2}+ bx + c$

Before the focus can be found, it is necessary to find the vertex (h, k) . This is located at the point with abscissa

$x = -\frac{b}{2a} = -\frac{4}{2(4)}= -\frac{1}{2}$ .

Substitute this for to find the ordinate:

$y= 4x^{2} + 4x+ 8$

$y= 4 \left ( -\frac{1}{2} \right )^{2} + 4 \left ( -\frac{1}{2} \right )+ 8$

$= 4 \left ( \frac{1}{4} \right ) + 4 \left ( -\frac{1}{2} \right )+ 8$

= 1 - 2 + 8

= 7

The vertex of the parabola is $\left ( -\frac{1}{2} , 7 \right )$ .

The focus of a vertical parabola is located at the point

$\left ( h, k + \frac{1}{4a} \right )$ .

Setting $h = -\frac{1}{2} , k = 7, a = 4$ , the point has coordinates

$\left ( -\frac{1}{2}, 7 + \frac{1}{4 (4)} \right )$ .

$7 + \frac{1}{4 (4)} = 7 + \frac{1}{16} = \frac{113}{6}$ , so the focus is at

$\left ( -\frac{1}{2}, \frac{113}{16} \right )$ .

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Example Question #6 : Parabolas

Give the equation of the directrix of the parabola of the equation

$x=2 y^{2}+ 4y - 8$

Possible Answers:

$x = - \frac{81}{8}$

$x = - \frac{7}{8}$

$x = \frac{1}{8}$

$x = - \frac{79}{8}$

$x = - \frac{9}{8}$

Correct answer:

$x = - \frac{81}{8}$

Explanation:

The parabola in question is a horizontal parabola. Its equation is in the standard form

$y = ax^{2}+ bx + c$

Before the directrix can be found, it is necessary to find the vertex (h, k) . This is located at the point with ordinate

$y= -\frac{b}{2a} = - \frac{4}{2(2)} = -1$

and abscissa

$x=2 (-1)^{2}+ 4(-1) - 8$

= 2(1)+ 4(-1)- 8

= 2-4 -8

= -10

That is, the vertex is at (-10, -1) .

The directrix of the parabola is the line of the equation $x = h - \frac{1}{4a}$ , which is

$x = -10- \frac{1}{4(2)}$

$x = -10- \frac{1}{8}$

$x = - \frac{81}{8}$

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College Algebra : Parabolas

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #1 : Parabolas

Example Question #2 : Parabolas

Example Question #3 : Parabolas

Example Question #2 : Parabolas

Example Question #2 : Parabolas

Example Question #2 : Parabolas

Example Question #3 : Parabolas

Example Question #25 : Graphs

Example Question #5 : Parabolas

Example Question #6 : Parabolas

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