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HiSET: Math : HiSet: High School Equivalency Test: Math

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125 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Identification

The equation

$x^{2} + 34 = 17x$

has two distinct solutions. What is their sum?

Possible Answers:

Correct answer:

Explanation:

It is not necessary to actually find the solutions to a quadratic equation to determine the sum of its solutions.

First, get the equation in standard form $ax^{2}+ bx + c = 0$ by subtracting 17x from both sides:

$x^{2} + 34 = 17x$

$x^{2} + 34 -17x = 17x - 17x$

$x^{2} -17x + 34 = 0$

If a quadratic equation has two distinct solutions, which we are given here, their sum is the linear coefficient . In this problem, b= -17 , making the correct choice.

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Example Question #271 : Hi Set: High School Equivalency Test: Math

The graph of the polynomial function

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

has one and only one zero on the interval (0, 1) . On which subinterval is it located?

Possible Answers:

(0.8, 1)

(0.4, 0.6)

(0,6, 0.8)

(0.2, 0.4)

(0, 0.2)

Correct answer:

(0.4, 0.6)

Explanation:

The Intermediate Value Theorem (IVT) states that if the graph of a function is continuous on an interval [a, b] , and f(a) and f(b) differ in sign, then has a zero on (a,b) . Consequently, the way to answer this question is to determine the signs of on the endpoints of the subintervals - $\left \{ 0, 0.2, 0.4, 0.6, 0.8, 1 \right \}$ . We can do this by substituting each value for as follows:

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

$f(0)= 0^{3}+0^{2}-5(0)+2 = 0+0-0+2= 2$

$f(0.2)= 0.2^{3}+0.2^{2}-5(0.2)+2 = 0.008+0.04-1+2 = 1.048$

$f(0.4)= 0.4^{3}+0.4^{2}-5(0.4)+2= 0.064+ 0.16- 2+2 = 0.224$

$f(0.6)= 0.6^{3}+0.6^{2}-5(0.6)+2= 0.216+ 0.36- 3+2 = -0.424$

$f(0.8)= 0.8^{3}+0.8^{2}-5(0.8)+2= 0.512+ 0.64- 4+2 = -0.848$

$f(1)=1^{3}+1^{2}-5(1)+2 =1+1-5+2=-1$

assumes positive values for x= 0, 0.2, 0.4 and negative values for x = 0.6, 0.8, 1 . By the IVT, has a zero on (0.4, 0.6) .

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Example Question #1 : Use The Zeros To Construct A Rough Graph Of A Function

The graph of a function is shown below, with labels on the y-axis hidden.

Graph zeroes 2

Determine which of the following functions best fits the graph above.

Possible Answers:

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

$f(x)=\frac{x+1}{x+2}$

f(x)=-2(x+1)(x-2)(x+2)

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

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

Correct answer:

f(x)=-2(x+1)(x-2)(x+2)

Explanation:

Use the zeroes of the graph to determine the matching function. Zeroes are values of x where f(x)=0 . In other words, they are points on the graph where the curve touches zero.

Visually, you can see that the curve crosses the x-axis when x=-2 , x=-1 , and x=2 . Therefore, you need to look for a function that will equal zero at these x values.

A function with a factor of (x+2) will equal zero when x=-2 , because the factor of (x+2) will equal zero. The matching factors for the other two zeroes, x=-1 and x=2 , are (x+1) and (x-2) , respectively.

The answer choice f(x)=-2x^2(x+1)(x-2)(x+2) has all of these factors, but it is not the answer because it has an additional zero that would be visible on the graph. Notice it has a factor of x^2 , which results in a zero at x=0 . This additional zero that isn't present in the graph indicates that this cannot be matching function.

f(x)=-2(x+1)(x-2)(x+2) is the answer because it has all of the required factors and, as a result, the required zeroes, while not having additional zeroes. Notice that the constant coefficient of negative 2 does not affect where the zeroes are.

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Example Question #1 : Use The Zeros To Construct A Rough Graph Of A Function

Graph zeroes

Which of the functions below best matches the graphed function?

Possible Answers:

f(x)= x(3x-5)(x+2)

f(x)= x(3x+5)(x-2)

f(x)= (3x+5)(x+2)

f(x)= x(3x-5)(x-2)

f(x)= x(x+2)^2

Correct answer:

f(x)= x(3x-5)(x+2)

Explanation:

First, look at the zeroes of the graph. Zeroes are where the function touches the x-axis (i.e. values of where f(x)=0 ).

The graph shows the function touching the x-axis when x=-2 , x=0 , and at a value in between 1.5 and 2.

Notice all of the possible answers are already factored. Therefore, look for one with a factor of (which will make f(x)=0 when x=0 ), a factor of (x+2) to make f(x)=0 when x=-2 , and a factor which will make f(x)=0 when is at a value between 1.5 and 2.

f(x)= x(3x-5)(x+2)

This function fills the criteria; it has an and an (x+2) factor. Additionally, the third factor, (3x-5) , will result in f(x)=0 when $x=\frac{5}{3}$ , which fits the image. It also does not have any extra zeroes that would contradict the graph.

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Example Question #1 : Rate Of Change Of A Function

What is the slope of the line given by the following table?

$\begin{tabular}{ | c | c |} \hline x & y \\ \hline 0 & 2 \\ \hline 1 & 3 \\ \hline 2 & 4 \\ \hline \end{tabular}$

Possible Answers:

$\frac{2}{3}$

$-\frac{1}{3}$

$\frac{1}{2}$

Correct answer:

Explanation:

Given two points

$(x_{0},y_{0})$ and $(x_{1},y_{1})$

the formula for a slope is

$m=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}$ .

Thus, since our given table is

$\begin{tabular}{ | c | c |} \hline x & y \\ \hline 0 & 2 \\ \hline 1 & 3 \\ \hline 2 & 4 \\ \hline \end{tabular}$

we select two points, say

(0,2) and (1,3)

and use the slope formula to compute the slop.

Thus,

$m = \frac{3-2}{1-0}=\frac{1}{1} = 1$ .

Hence, the slope of the line generated by the table is

m=1 .

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Example Question #1 : Rate Of Change Of A Function

Define $f(x) = x^{2}+ x$ .

Give the average rate of change of over the interval [2,3] .

Possible Answers:

7.2

3.6

Correct answer:

Explanation:

The average rate of change of a function over an interval [a, b] is equal to

$\frac{f(b)-f(a)}{b-a}$

Setting a = 2, b= 3 , this is

$\frac{f(3)-f(2)}{3-2} = \frac{f(3)-f(2)}{1} = f(3)-f(2)$

Evaluate f(3) and f(2) by substitution:

$f(x) = x^{2}+ x$

$f(3) = 3^{2}+ 3= 9+ 3 = 12$

$f(2) = 2^{2}+ 2 = 4+ 2 = 6$

$\frac{f(3)-f(2)}{3-2} = 12 - 6 = 6$ ,

the correct response.

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Example Question #1 : Calculate And Interpret Rate Of Change Over A Specified Interval

Define $f(x) = \left\{\begin{matrix} x^{2}+x ,&x \le 0 \\ x^{2}-x, & x> 0\end{matrix}\right.$ .

Give the average rate of change of over the interval [-1, 1] .

Possible Answers:

$-\frac{1}{2}$

$\frac{1}{2}$

Correct answer:

Explanation:

The average rate of change of a function over an interval [a, b] is equal to

$\frac{f(b)-f(a)}{b-a}$

Setting a = -1, b= 1 , this is

$\frac{f(1)-f(-1)}{1-(-1)} = \frac{f(1)-f(-1)}{2}$

Evaluate f(1) using the definition of for x>0 :

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

$f(1) = 1^{2}-1 = 1 - 1 = 0$

Evaluate f(-1) using the definition of for $x\le 0$ :

$f(x) = x^{2}+x$

$f(-1) = (-)1^{2}+ (-1) = 1 - 1 = 0$

The average rate of change is therefore

$\frac{f(1)-f(-1)}{2} = \frac{0-0}{2} = 0$ .

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Example Question #1 : Estimate Rate Of Change From A Graph

Estimate slope

The graph of a function is given above, with the coordinates of two points on the curve shown. Use the coordinates given to approximate the rate of change of the function between the two points.

Possible Answers:

$\frac{3}{5}$

$-\frac{11}{7}$

$\frac{7}{11}$

$\frac{5}{3}$

$-\frac{3}{5}$

Correct answer:

$\frac{3}{5}$

Explanation:

The rate of change between two points on a curve can be approximated by calculating the change between two points.

Let (x_1,y_1) be the coordinates of the first point and (x_2,y_2) be the coordinates of the second point. Then the formula giving approximate rate of change is:

$\frac{y_2 - y_1} {x_2 - x_1}$

Notice that the numerator is the overall change in y, and the denominator is the overall change in x.

The calculation for the problem proceeds as follows:

Let (3,2) be the first point and (8,5) be the second point. Substitute in the values from these coordinates:

$\frac{y_2 - y_1} {x_2 - x_1} \rightarrow \frac {5-2}{8-3}$

Subtract to get the final answer:

$\frac{3}{5}$

Note that it does not matter which you assign to be the first point and which you assign to be the second, as it will lead to the same value due to negatives canceling out.

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Example Question #1 : Estimate Rate Of Change From A Graph

Graph 1

Above is the graph of a function . Estimate the rate of change of on the interval [-2, -1.5]

Possible Answers:

Correct answer:

Explanation:

The rate of change of a function on the interval [a, b] is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Set a = -2, b = -1.5 . Refer to the graph of the function below:

Graph 1

The graph passes through (-2, 2) and (-1.5, 5) .

f(-2) = 2 , f(-1.5)= 5 . Thus,

$r = \frac{f(-1.5)-f(-2)}{-1.5 - (-2)} = \frac{5-2}{0.5} = \frac{3}{0.5} = 6$ ,

the correct response.

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Example Question #3 : Rate Of Change Of A Function

Graph 1

Above is the graph of a function , which is defined and continuous on $(-\infty, \infty)$ . The average rate of change of on the interval [1, 5] is 4. Estimate f(5) .

Possible Answers:

Correct answer:

Explanation:

The rate of change of a function on the interval [a, b] is equal to

$r = \frac{f(b)-f(a)}{b - a}$ .

Set a = 1, b = 5, r =4 . Examine the figure below:

Graph 1

The graph passes through the point (1, 2) , so f(1) = 2 . Therefore,

$4 = \frac{f(5)-f(1)}{5-1}$

and, substituting,

$4 = \frac{f(5)-2}{4}$

Solve for f(5) using algebra:

$\frac{f(5)-2}{4} \cdot 4 = 4 \cdot 4$

f(5)-2 =16

f(5)-2+ 2 =16 + 2

f(5)= 18 ,

the correct response.

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HiSET: Math : HiSet: High School Equivalency Test: Math

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #1 : Identification

Example Question #271 : Hi Set: High School Equivalency Test: Math

Example Question #1 : Use The Zeros To Construct A Rough Graph Of A Function

Example Question #1 : Use The Zeros To Construct A Rough Graph Of A Function

Example Question #1 : Rate Of Change Of A Function

Example Question #1 : Rate Of Change Of A Function

Example Question #1 : Calculate And Interpret Rate Of Change Over A Specified Interval

Example Question #1 : Estimate Rate Of Change From A Graph

Example Question #1 : Estimate Rate Of Change From A Graph

Example Question #3 : Rate Of Change Of A Function

All HiSET: Math Resources

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