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Algebra II : Functions and Graphs

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Example Questions

Example Question #11 : Parabolic Functions

Find the location of the vertex for the parabola. Is it a max or min?

y=-2x^2+6x+7

Possible Answers:

$\textup{Min at }x=\frac{2}{3}$

$\textup{Max at }y=\frac{3}{2}$

$\textup{Max at }x=\frac{2}{3}$

$\textup{Max at }x=\frac{3}{2}$

$\textup{Min at }x=\frac{3}{2}$

Correct answer:

$\textup{Max at }x=\frac{3}{2}$

Explanation:

The polynomial is written in the form of: y=ax^2+bx+c

This is the standard form for a parabola.

Write the vertex formula, and substitute the known values:

$x=-\frac{b}{2a}=-\frac{6}{2(-2)} = \frac{6}{4} =\frac{3}{2}$

The vertex is at: $x=\frac{3}{2}$

Since the coefficient of is negative, the curve will open downward, and will have a maximum.

The answer is: $\textup{Max at }x=\frac{3}{2}$

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Example Question #411 : Functions And Graphs

A particular parabola has it's vertex at (-5,-5) , and an x-intercept at the origin. Determine the equation of the parabola.

Possible Answers:

None of these

.2x^2+2x=y

x^2+10x=y

x^2+2x=y

x^2+5x=y

Correct answer:

.2x^2+2x=y

Explanation:

General parabola equation:

ax^2+bx+c=y

Vertex formula:

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

Where is the value at the vertex.

Combining equations:

ax^2-2ahx+c=y

Plugging in values for vertex:

a(-5)^2-2a(-5)(-5)+c=(-5)

Solving for :

c=25a-5

Returning to:

ax^2+bx+c=y

combining equations:

ax^2+bx+25a-5=y

Plugging in values of given intercept:

a*0^2+b*0+25a-5=0

Solving for

a=.2

Plugging in value:

c=25(.2)-5

c=0

Plugging in values for the vertex:

.2(-5)^2+b(-5)+0=(-5)

5-5b=-5

b=2

Final equation:

.2x^2+2x=y

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Example Question #411 : Functions And Graphs

Which of the following functions represents a parabola?

Possible Answers:

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

$f(x) = \frac{4x^{2}}{x^{3}}$

f(x) = x

$f(x) = x^{2} - y^{2}$

f(x) = 5

Correct answer:

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

Explanation:

A parabola is a curve that can be represented by a quadratic equation. The only quadratic here is represented by the function $f(x) = x^{2} - 9$ , while the others represent straight lines, circles, and other curves.

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Example Question #14 : Quadratic Functions

What is the point of the vertex of the parabola y=3(3-2x)^2-1 ? Is it a maximum or minimum?

Possible Answers:

$( \frac{3}{2},-1), \textup{Maximum}$

$( \frac{3}{2},-1), \textup{Minimum}$

$( \frac{2}{3},\frac{22}{3}), \textup{Maximum}$

$( \frac{2}{3},\frac{22}{3}), \textup{Minimum}$

$( -\frac{3}{2},107), \textup{Maximum}$

Correct answer:

$( \frac{3}{2},-1), \textup{Minimum}$

Explanation:

It is not necessary to FOIL the binomial in order to solve for the vertex. Switch the terms of the quantity (3-2x) , and this equation will be in vertex form:

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

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

Set the inner quantity equal to zero.

-2x+3 = 0

Subtract three on both sides.

-2x+3 -3= 0-3

-2x=-3

Divide by negative two on both sides.

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

The location of the vertex is at $x = \frac{3}{2}$ .

To determine the point, substitute the value $x = \frac{3}{2}$ back to the original equation.

$y=3(3-2(\frac{3}{2}))^2-1 = 3(3-3)^2-1 = 0-1 = -1$

The point of the vertex is at: $( \frac{3}{2},-1)$

Since this parabola opens upward, the point of the vertex will be a minimum.

The answer is: $( \frac{3}{2},-1), \textup{Minimum}$

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Example Question #932 : Algebra Ii

Where is the vertex located for the given function? y=-2x^2-x-1

Possible Answers:

$(-\frac{1}{4},-\frac{5}{8})$

$(-\frac{1}{4},-\frac{7}{8})$

$(-\frac{1}{4},\frac{5}{8})$

$(-\frac{1}{4},-\frac{11}{8})$

$(\frac{1}{4},-\frac{11}{8})$

Correct answer:

$(-\frac{1}{4},-\frac{7}{8})$

Explanation:

Write the vertex formula.

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

The given equation is already in standard polynomial form.

a=-2, b=-1, c=-1

Substitute the known values into the formula.

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

Substitute this value back into the original equation to determine the y value.

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

Simplify this expression.

$y=-2(\frac{1}{16})+\frac{1}{4}-1 = -\frac{1}{8}+\frac{1}{4}-1$

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

The vertex is located at: $(-\frac{1}{4},-\frac{7}{8})$

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Example Question #11 : Parabolic Functions

Which of these functions represent a parabola?

Possible Answers:

y=5x-4

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

0=3x^2+7-y

y^2=x^2

Correct answer:

0=3x^2+7-y

Explanation:

A parabola is a curve that is represented by a quadratic function. In this case, the only answer that qualifies is 0=3x^2+7-y . The other answers represent straight lines, and other types of curves.

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Example Question #12 : Quadratic Functions

Where is the vertex located for y=5x-3x^2 ?

Possible Answers:

$(\frac{3}{10}, \frac{123}{100})$

$(\frac{5}{6},\frac{25}{12} )$

$(\frac{5}{6},\frac{75}{4} )$

$(-\frac{5}{6},-\frac{25}{4} )$

$(-\frac{5}{6},-\frac{75}{4} )$

Correct answer:

$(\frac{5}{6},\frac{25}{12} )$

Explanation:

Rewrite the equation in standard polynomial form.

y=ax^2+bx+c

y=-3x^2+5x

a=-3, b=5, c=0

Write the vertex formula and substitute the known coefficients.

$x=-\frac{b}{2a} = -\frac{5}{2(-3)} = \frac{5}{6}$

The x-value of the vertex is $\frac{5}{6}$ .

Substitute this value back to the original equation.

$y=-3(\frac{5}{6})^2+5(\frac{5}{6})$

$y=-3(\frac{5}{6})(\frac{5}{6})+5(\frac{5}{6})$

$y=-\frac{25}{12}+\frac{25}{6} = -\frac{25}{12}+\frac{25(2)}{6(2)} =-\frac{25}{12}+\frac{50}{12}$

$y=\frac{25}{12}$

The vertex is located at: $(\frac{5}{6},\frac{25}{12} )$

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Example Question #13 : Quadratic Functions

Determine the vertex given the function: y=3x^2-6

Possible Answers:

$\left(-\frac{1}{2}, \frac{21}{4}\right)$

(2,6)

(-2,6)

$\left(\frac{1}{2}, \frac{21}{4}\right)$

(0,-6)

Correct answer:

(0,-6)

Explanation:

The parabola is provided in the form of y=ax^2+bx+c .

Notice that the variable in this equation is zero.

This means that:

$x=-\frac{b}{2a} = 0$

Substitute this value back into the original equation y=3x^2-6 to determine the y-value.

y=3(0)^2-6

y=-6

The vertex is located at: (0,-6)

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Example Question #3 : Graphing Quadratic Functions

All of the following are equations of down-facing parabolas EXCEPT:

Possible Answers:

$y = -x^{2}$

$y = -6(x-4)^{2} -6$

$y = -(x+3)^{2}$

$y = -6(x+4)^{2} -6$

$x = 2(y+3)^{2} -5$

Correct answer:

$x = 2(y+3)^{2} -5$

Explanation:

A parabola that opens downward has the general formula

$y = -(x-a)^{2} + b$ ,

as the negative sign in front of the (x-a) term makes flips the parabola about the horizontal axis.

By contrast, a parabola of the form $x = y^{2}$ rotates about the vertical axis, not the horizontal axis.

Therefore, $x = 2(y+3)^{2} -5$ is not the equation for a parabola that opens downward.

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Example Question #4 : Graphing Quadratic Functions

Consider the equation:

$y = 3x^{2} - 10x +5$

The vertex of this parabolic function would be located at:

Possible Answers:

$(\frac{5}{3}, \frac{-10}{3})$

$(\frac{5}{3}, \frac{2}{25})$

$(\frac{-5}{3}, 30)$

$(\frac{-5}{3}, \frac{-10}{3})$

$(\frac{5}{3}, \frac{27}{25})$

Correct answer:

$(\frac{5}{3}, \frac{-10}{3})$

Explanation:

For any parabola, the general equation is

$y = ax^{2} + bx +c$ , and the x-coordinate of its vertex is given by

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

For the given problem, the x-coordinate is

$x = \frac{-(-10)}{2(3)} = \frac{5}{3}$ .

To find the y-coordinate, plug $\frac{5}{3}$ into the original equation:

$y = 3(\frac{5}{3})^{2} - 10(\frac{5}{3}) +5 = \frac{-10}{3}$

Therefore the vertex is at $(\frac{5}{3}, \frac{-10}{3})$ .

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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 #11 : Parabolic Functions

Example Question #411 : Functions And Graphs

Example Question #411 : Functions And Graphs

Example Question #14 : Quadratic Functions

Example Question #932 : Algebra Ii

Example Question #11 : Parabolic Functions

Example Question #12 : Quadratic Functions

Example Question #13 : Quadratic Functions

Example Question #3 : Graphing Quadratic Functions

Example Question #4 : Graphing Quadratic Functions

All Algebra II Resources

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