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Algebra II : Functions and Graphs

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Example Questions

Example Question #401 : Functions And Graphs

What are the -intercepts of the equation?

$y=\frac{x^2-9}{x^2-16}$

Possible Answers:

x=3,-3

There are no -intercepts.

x=3

x=2

x=4,-4

Correct answer:

x=3,-3

Explanation:

To find the x-intercepts of the equation, we set the numerator equal to zero.

0=x^2-9

9=x^2

$\sqrt{9}=\sqrt{x^2}$

3,-3=x

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Example Question #2 : Parabolic Functions

What is the minimum possible value of the expression below?

$3x^{2}+6x-10$

Possible Answers:

The expression has no minimum value.

Correct answer:

Explanation:

We can determine the lowest possible value of the expression by finding the -coordinate of the vertex of the parabola graphed from the equation $y=3x^{2}+6x-10$ . This is done by rewriting the equation in vertex form.

$3x^{2}+6x-10$

$3(x^{2}+2x)-10$

$3(x^{2}+2x+1)-3* 1-10$

$3(x+1)^{2}-13$

The vertex of the parabola $y=3(x+1)^{2}-13$ is the point (-1, -13) .

The parabola is concave upward (its quadratic coefficient is positive), so represents the minimum value of . This is our answer.

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Example Question #924 : Algebra Ii

Find the coordinates of the vertex of this quadratic function:

$y=-2x^{2}+6x-5$

Possible Answers:

$\left(\frac{3}{2},-\frac{1}{2}\right)$

$\left(-\frac{3}{2}, -\frac{1}{2}\right)$

$\left(\frac{2}{3}, -\frac{1}{2}\right)$

$\left(-\frac{2}{3}, -\frac{1}{2}\right)$

$\left(\frac{3}{2}, \frac{1}{2}\right)$

Correct answer:

$\left(\frac{3}{2},-\frac{1}{2}\right)$

Explanation:

Vertex of quadratic equation $y=ax^{2}+bx+c$ is given by (h,k) .

$h=-\frac{b}{2a}, k=c-\frac{b^{2}}{4a}$

For $y=-2x^{2}+6x-5$ ,

$h=-\frac{6}{2\times (-2)}=\frac{3}{2},$

$k=-5-\frac{6^{2}}{4\times (-2)}=-5+\frac{9}{2}=-\frac{1}{2}$ ,

so the coordinate of vertex is $\left(\frac{3}{2}, -\frac{1}{2}\right)$ .

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Example Question #2 : Parabolic Functions

What are the x-intercepts of the graph of $y=8x^{2}+2x-3$ ?

Possible Answers:

$-\frac{1}{2}\ and -\frac{3}{4}$

$-\frac{1}{2}\ and\ \frac{3}{4}$

$\frac{1}{2} \ and -\frac{3}{4}$

$-\frac{3}{2}\ and\ -\frac{1}{4}$

$-\frac{3}{2}\ and\ \frac{1}{4}$

Correct answer:

$\frac{1}{2} \ and -\frac{3}{4}$

Explanation:

Assume y=0,

$y=8x^{2}+2x-3=0$

(2x-1)(4x+3)=0

$x=\frac{1}{2}$ , $x=-\frac{3}{4}$

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Example Question #1 : Parabolic Functions

Find the vertex of the parabola given by the following equation:

f(x)=-2x^2-12x-23

Possible Answers:

(4,1)

(1,-7)

(-4,12)

(2,4)

(-3,-5)

Correct answer:

(-3,-5)

Explanation:

In order to find the vertex of a parabola, our first step is to find the x-coordinate of its center. If the equation of a parabola has the following form:

f(x)=ax^2+bx+c

Then the x-coordinate of its center is given by the following formula:

$x_{center}=-\frac{b}{2a}$

For the parabola described in the problem, a=-2 and b=-12, so our center is at:

$x_{center}=-\frac{-12}{2(-2)}=-3$

Now that we know the x-coordinate of the parabola's center, we can simply plug this value into the function to find the y-coordinate of the vertex:

f(-3)=-2(-3)^2-12(-3)-23=-5

So the vertex of the parabola given in the problem is at the point (-3,-5)

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Example Question #2 : Parabolic Functions

Give the minimum value of the function $f(x) = 2 x^{2} - 5x + 9$ .

Possible Answers:

$18\frac{3}{8}$

This function does not have a minimum.

$5\frac{7}{8}$

Correct answer:

$5\frac{7}{8}$

Explanation:

This is a quadratic function. The -coordinate of the vertex of the parabola can be determined using the formula $x = - \frac{b}{2a}$ , setting a = 2,b = -5 :

$x = - \frac{b}{2a} = - \frac{-5}{2 \cdot 2} =\frac{5}{4}$

Now evaluate the function at $x = \frac{5}{4}$ :

$f(x) = 2 x^{2} - 5x + 9$

$f \left ( \frac{5}{4} \right ) = 2 \left ( \frac{5}{4} \right )^{2} - 5 \left ( \frac{5}{4} \right )+ 9$

$= 2 \left (\frac{25}{16} \right ) - 5 \left ( \frac{5}{4} \right )+ 9$

$= \frac{25}{8} - \frac{25}{4} + 9$

$= \frac{25}{8} - \frac{50}{8} + \frac{72}{8} =\frac{47}{8} =5\frac{7}{8}$

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Example Question #1 : Understand Linear And Nonlinear Functions: Ccss.Math.Content.8.F.A.3

What is the equation of a parabola with vertex (4,1) and -intercept (0,-7) ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

From the vertex, we know that the equation of the parabola will take the form $y = a (x - 4) ^{2}+ 1$ for some .

To calculate that , we plug in the values from the other point we are given, x = 0, y=-7 , and solve for :

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

$-7= a \cdot (0 - 4)^{2} + 1$

$-7= a \cdot (- 4)^{2} + 1$

-7= 16a + 1

16a = -8

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

Now the equation is $y = -\frac{1}{2} (x - 4)^{2} + 1$ . This is not an answer choice, so we need to rewrite it in some way.

Expand the squared term:

$y = -\frac{1}{2} (x^{2} -8x + 16) + 1$

Distribute the fraction through the parentheses:

$y = -\frac{1}{2} x^{2} +4x - 8 + 1$

Combine like terms:

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

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Example Question #7 : Quadratic Functions

Determine the maximum or minimum of y=-5x^2+25x+5 .

Possible Answers:

$\textup{Max at: }x=\frac{5}{2}$

$\textup{Max at: }x=-\frac{2}{5}$

$\textup{Min at: }x=-\frac{5}{2}$

$\textup{Max at: }x=-\frac{5}{2}$

$\textup{Min at: }x=\frac{5}{2}$

Correct answer:

$\textup{Max at: }x=\frac{5}{2}$

Explanation:

To find the max or min of y=-5x^2+25x+5 , use the vertex formula and substitute the appropriate coefficients.

$x=\frac{-b}{2a}=\frac{-25}{2(-5)}= \frac{25}{10}=\frac{5}{2}$

Since the leading coefficient of -5x^2 is negative, the parabola opens down, which indicates that there will be a maximum.

The answer is: $\textup{Max at: }x=\frac{5}{2}$

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Example Question #3 : Parabolic Functions

Factorize: 6x^2-17x+10

Possible Answers:

(2x-5)(3x-2)

(6x-5)(x-2)

(2x-10)(3x-1)

(6x-5)(x+2)

-(6x+5)(x+2)

Correct answer:

(6x-5)(x-2)

Explanation:

To simplify 6x^2-17x+10 , determine the factors of the first and last term.

The factor possibilities of 6x^2 :

x, 2x, 3x, 6x

The factor possibilities of :

1,2,5,10

Determine the signs. Since there is a positive ending term and a negative middle term, the signs of the binomials must be both negative. Write the pair of parenthesis.

(a -b)(c-d) = ac-ad-bc-bd

These factors must be manipulated by trial and error to determine the middle term.

The correct selection is x, 6x,2, 5 .

The answer is (6x-5)(x-2) .

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Example Question #403 : Functions And Graphs

Find the location of the vertex of the parabola: y=2(3x+3x^2)

Possible Answers:

$x=-\frac{3}{2}$

x=-1

$x=-\frac{1}{4}$

$x=-\frac{1}{2}$

$x=-\frac{1}{3}$

Correct answer:

$x=-\frac{1}{2}$

Explanation:

Multiply the two through the binomial.

y=2(3x+3x^2) = 6x+6x^2 = 6x^2+6x

Now that this is in the order of the polynomial y=ax^2+bx+c , we can use the vertex formula.

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

Substitute the known coefficients.

$x=-\frac{b}{2a} =-\frac{6}{2(6)}=-\frac{1}{2}$

The answer is: $x=-\frac{1}{2}$

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Algebra II : Functions and Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #401 : Functions And Graphs

Example Question #2 : Parabolic Functions

Example Question #924 : Algebra Ii

Example Question #2 : Parabolic Functions

Example Question #1 : Parabolic Functions

Example Question #2 : Parabolic Functions

Example Question #1 : Understand Linear And Nonlinear Functions: Ccss.Math.Content.8.F.A.3

Example Question #7 : Quadratic Functions

Example Question #3 : Parabolic Functions

Example Question #403 : Functions And Graphs

All Algebra II Resources

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