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Algebra 1 : Functions and Lines

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Example Question #2 : Introduction To Functions

Which graph depicts a function?

Possible Answers:

Correct answer:

Explanation:

A function may only have one y-value for each x-value.

The vertical line test can be used to identify the function. If at any point on the graph, a straight vertical line intersects the curve at more than one point, the curve is not a function.

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Example Question #71 : How To Find The Solution For A System Of Equations

The graph below is the graph of a piece-wise function in some interval. Identify, in interval notation, the decreasing interval.

Possible Answers:

$\left ( 1,3 \right )$

$(-\infty ,-2]\cup (-\infty ,1)\cup (1,3)$

$\left ( -\infty ,-2\right )\cup \left ( 1,3\right]$

$\left ( -\infty ,\infty \right )$

$\left ( -\infty ,-2\right ]\cup \left ( 1,3 \right )$

Correct answer:

$\left ( 1,3 \right )$

Explanation:

As is clear from the graph, in the interval between ( included) to , the f(x) is constant at and then from x=1 ( not included) to x=3 ( not included), the f(x) is a decreasing function.

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Example Question #1 : Graphing Polynomial Functions

Which equation best represents the following graph?

Graph6

Possible Answers:

None of these

y(x)=-x-1

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

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

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

Correct answer:

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

Explanation:

We have the following answer choices.

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

The first equation is a cubic function, which produces a function similar to the graph. The second equation is quadratic and thus, a parabola. The graph does not look like a prabola, so the 2nd equation will be incorrect. The third equation describes a line, but the graph is not linear; the third equation is incorrect. The fourth equation is incorrect because it is an exponential, and the graph is not an exponential. So that leaves the first equation as the best possible choice.

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Example Question #1 : Solving And Graphing Exponential Equations

What is the horizontal asymptote of the graph of the equation $y = -2e^{x-3} -5$ ?

Possible Answers:

y=-2

$y=-\frac{5}{2}$

y=3

y=5

y=-5

Correct answer:

y=-5

Explanation:

The asymptote of this equation can be found by observing that $e^{x-3} > 0$ regardless of . We are thus solving for the value of as $e^{x-3}$ approaches zero.

$-2e^{x-3} < -2\cdot 0$

$-2e^{x-3} < 0$

$-2e^{x-3} -5 < 0-5$

$-2e^{x-3} -5 < -5$

y<-5

So the value that cannot exceed is , and the line y = -5 is the asymptote.

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Example Question #2 : Solving And Graphing Exponential Equations

What is/are the asymptote(s) of the graph of the function

$f(x) = 5e^{x-2}-2$ ?

Possible Answers:

y = -2

y = -2, x=2

x=2

x=-2

y = -2, x=-2

Correct answer:

y = -2

Explanation:

An exponential equation of the form $f(x) = ae^{x-h} +k$ has only one asymptote - a horizontal one at y = k . In the given function, k = -2 , so its one and only asymptote is y = -2 .

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Example Question #1 : How To Graph An Exponential Function

$\textup{What is the \textit{x}-intercept of the equation } y=e^{x-1}-1?$

Possible Answers:

$\textup{The graph does not intercept the \textit{x}-axis.}$

$\textup{None of the above.}$

$-\infty$

Correct answer:

Explanation:

$\\ \\ \textup{Solve the equation } \\0=e^{x-1} -1\\ e^{x-1}=1\\ {x-1}=\textup{ln}(1)=0\\ x=1.\\\textup{ Remember }e^0=1 \textup{, so when plugging } x=1 \textup{ in the initial equation, you obtain }\\ e^{1-1}-1=e^0-1=1-1=0.$

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Example Question #1 : How To Graph A Two Step Inequality

Which graph depicts the following inequality?

$x\geq4y+12$

Possible Answers:

Question_12_incorrect_2

Question_12_incorrect_3

No real solution.

Question_12_correct

Question_12_incorrect_1

Correct answer:

Question_12_correct

Explanation:

Let's put the inequality in slope-intercept form to make it easier to graph:

$x\geq4y+12$

$x-12\geq4y+12-12$

$x-12\geq4y$

$\frac{x-12}{4}\geq\frac{4y}{y}$

$\frac{1}{4}x-3\geq y$

The inequality is now in slope-intercept form. Graph a line with slope $\frac{1}{4}$ and y-intercept .

Because the inequality sign is greater than or equal to, a solid line should be used.

Next, test a point. The origin (0,0) is good choice. Determine if the following statement is true:

$\frac{1}{4}(0)-3\geq0$

The statement is false. Therefore, the section of the graph that does not contain the origin should be shaded.

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Example Question #1 : 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 #1 : Graphing Quadratic Functions

What is the vertex of the function $\small f(x)=2(x-4)^2+7$ ? Is it a maximum or minimum?

Possible Answers:

$\small (-4,7)$ ; maximum

$\small (4,7)$ ; minimum

$\small (4,7)$ ; maximum

$\small (-4,7)$ ; minimum

Correct answer:

$\small (4,7)$ ; minimum

Explanation:

The equation of a parabola can be written in vertex form: $\small f(x)=a(x-h)^2+k$ .

The point $\small (h,k)$ in this format is the vertex. If $\small a$ is a postive number the vertex is a minimum, and if $\small a$ is a negative number the vertex is a maximum.

$\small f(x)=2(x-4)^2+7$

In this example, $\small a=2$ . The positive value means the vertex is a minimum.

$\small h=4\ \text{and}\ k=7$

$\small (h,k)=(4,7)$

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Example Question #1 : Graphing Polynomial Functions

Which of the graphs best represents the following function?

$f(x)=\frac{9}{10}x^2-7x+2$

Possible Answers:

Graph_cube_

Graph_line_

None of these

Graph_parabola_

Graph_exponential_

Correct answer:

Graph_parabola_

Explanation:

$f(x)=\frac{9}{10}x^2-7x+2$

The highest exponent of the variable term is two ( x^2 ). This tells that this function is quadratic, meaning that it is a parabola.

The graph below will be the answer, as it shows a parabolic curve.

Graph_parabola_

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Algebra 1 : Functions and Lines

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #2 : Introduction To Functions

Example Question #71 : How To Find The Solution For A System Of Equations

Example Question #1 : Graphing Polynomial Functions

Example Question #1 : Solving And Graphing Exponential Equations

Example Question #2 : Solving And Graphing Exponential Equations

Example Question #1 : How To Graph An Exponential Function

Example Question #1 : How To Graph A Two Step Inequality

Example Question #1 : Parabolic Functions

Example Question #1 : Graphing Quadratic Functions

Example Question #1 : Graphing Polynomial Functions

All Algebra 1 Resources

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