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HSPT Math : HSPT Mathematics

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Example Questions

Example Question #1475 : Concepts

Axes

Give the equation of the red line in slope-intercept form.

Possible Answers:

$y =- \frac{7}{4}x + 6$

$y = \frac{7}{4}x + 6$

$y = \frac{4}{7}x + 6$

$y =- \frac{7}{4}\left (x + 6 \right )$

$y = -\frac{4}{7}x + 6$

Correct answer:

$y = \frac{7}{4}x + 6$

Explanation:

The slope of the line is

$m = \frac{y_{2}- y_{1}}{x_{2}- x_{1}} = \frac{6- (-1)}{0-(-4)} = \frac{7}{4}$

The -intercept of the line has -coordinate b= 6

The slope-intercept form can be written:

y = mx+b

Replace:

$y = \frac{7}{4}x+6$

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Example Question #1476 : Concepts

Axes_2

The green and blue lines are perpendicular: What is the slope of the blue line?

Possible Answers:

It cannot be determined from the information given.

$-\frac{1}{3}$

$\frac{1}{3}$

Correct answer:

Explanation:

The slope of the blue line, being perpendicular to the green line, is the opposite of the reciprocal of the slope of the green line. The slope of the green line can be found using the slope formula:

$m = \frac{y_{2}- y_{1}}{x_{2}- x_{1}} = \frac{10-0}{0-30} = -\frac{1}{3}$

The opposite of the reciprocal of $-\frac{1}{3}$ is 3, and this is the slope of the blue line.

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Example Question #1477 : Concepts

A line segment on the coordinate plane has endpoints (a-5,-b) and (b+8, a-4) . In terms of and , as applicable, give the -coordinate of its midpoint.

Possible Answers:

a - b - 4

a + b + 3

$\frac{a + b + 3}{2}$

$\frac{a - b - 4}{2}$

$\frac{a - b - 13}{2}$

Correct answer:

$\frac{a + b + 3}{2}$

Explanation:

The -coordinate of the midpoint of a line segment is the mean of the -coordinates of its endpoints. Therefore, the -coordinate is

$x= \frac{(a-5)+(b+8)}{2} = \frac{a+b-5+8}{2} = \frac{a+b+3}{2}$ .

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Example Question #1478 : Concepts

A line segment on the coordinate plane has endpoints (a-5,-b) and (b+8, a-4) . In terms of and , as applicable, give the -coordinate of its midpoint.

Possible Answers:

a + b + 3

a - b - 4

$\frac{a - b - 4}{2}$

$\frac{a + b + 3}{2}$

$\frac{a + b - 4}{2}$

Correct answer:

$\frac{a - b - 4}{2}$

Explanation:

The -coordinate of the midpoint of a line segment is the mean of the -coordinates of its endpoints. Therefore, the -coordinate is

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

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Example Question #1 : How To Find The Endpoints Of A Line Segment

A line segment on the coordinate plane has midpoint $\left ( A + 5, B + 7\right )$ . One of its endpoints is $\left ( 10, 6 \right )$ . What is the -coordinate of the other endpoint, in terms of and/or ?

Possible Answers:

B + 8

2B + 20

2B + 1

B + 1

2B + 8

Correct answer:

2B + 8

Explanation:

Let be the -coordinate of the other endpoint. Since the -coordinate of the midpoint of the segment is the mean of those of the endpoints, we can set up an equation as follows:

$\frac{Y + 6 }{2} = B + 7$

$\frac{Y + 6 }{2} \cdot 2= \left ( B + 7 \right )\cdot 2$

Y + 6 = 2B + 14

Y + 6 -6 = 2B + 14 -6

Y = 2B + 8

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Example Question #2 : How To Find The Endpoints Of A Line Segment

A line segment on the coordinate plane has one endpoint at $\left ( 8, 10 \right )$ ; its midpoint is $\left ( A+B, A- B\right )$ . Which of the following gives the -coordinate of its other endpoint in terms of and ?

Possible Answers:

2 A + 2B+8

A + B- 16

2 A + 2B- 8

A + B- 8

A + B+8

Correct answer:

2 A + 2B- 8

Explanation:

To find the value of the -coordinate of the other endpoint, we will assign the variable . Then, since the -coordinate of the midpoint of the segment is the mean of those of its endpoints, the equation that we can set up is

$\frac{T + 8 }{2} = A + B$ .

We solve for :

$\frac{T + 8 }{2} \cdot 2 =\left ( A + B \right )\cdot 2$

T + 8 =2 A + 2B

T + 8 - 8 =2 A + 2B- 8

T =2 A + 2B- 8

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Example Question #1 : How To Find Out If Lines Are Perpendicular

Two perpendicular lines intersect at the point (5,4) . One line passes through point (3,1) ; the other passes through point $\left ( 4, Y\right )$ . Evaluate .

Possible Answers:

$Y = 2\frac{1}{2}$

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

$Y = 5\frac{1}{2}$

$Y = 4\frac{2}{3}$

$Y = 3\frac{1}{3}$

Correct answer:

$Y = 4\frac{2}{3}$

Explanation:

The line that passes through $\left ( 3,1\right )$ and $\left ( 5,4\right )$ has slope

$m = \frac{4-1}{5-3} = \frac{3}{2}$ .

The line that passes through $\left ( 4, Y\right )$ and (5,4) , being perpendicular to the first, has as its slope the opposite reciprocal of $\frac{3}{2}$ , or $-\frac{2}{3}$ .

Therefore, to find , we use the slope formula and solve for :

$-\frac{2}{3} = \frac{Y-4}{4-5}$

$-\frac{2}{3} = \frac{Y-4}{-1}$

$-\frac{2}{3} \cdot (-1) = \frac{Y-4}{-1} \cdot (-1)$

$\frac{2}{3}= Y-4$

$\frac{2}{3}+ 4= Y-4 + 4$

$4\frac{2}{3}= Y$

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Example Question #3 : Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

What is the -intercept of the graph of the function $f \left ( x\right ) = 2x^{2} - 7x + 5$ ?

Possible Answers:

$\left (0,2 \right )$

$\left (0,0 \right )$

$\left ( 0, -3 \frac{1}{2}\right )$

$\left (0,5 \right )$

$\left (0,-7 \right )$

Correct answer:

$\left (0,5 \right )$

Explanation:

The -intercept of the graph of a function is the point at which it intersects the -axis - that is, at which x = 0 . This point is $\left (0, f(0) \right )$ , so evaluate f(0) :

$f \left ( x\right ) = 2x^{2} - 7x + 5$

$f \left (0\right ) = 2 \cdot 0^{2} - 7 \cdot 0 + 5$

$f \left (0\right ) = 0- 0 + 5 = 5$

The -intercept is $\left (0,5 \right )$ .

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Example Question #2 : X And Y Intercept

Give the -intercept, if there is one, of the graph of the equation

$y = \frac{1}{x^{2}+ x} + 3$

Possible Answers:

$\left ( 0, \frac{3}{2}\right )$

(0,4)

The graph has no -intercept.

(0,2)

(0,3)

Correct answer:

The graph has no -intercept.

Explanation:

The -intercept is the point at which the graph crosses the -axis; at this point, the -coordinate is 0, so substitute for in the equation:

$y = \frac{1}{x^{2}+ x} + 3$

$y = \frac{1}{0^{2}+0} + 3$

$y = \frac{1}{0} + 3$

However, since this expression has 0 in a denominator, it is of undefined value. This means that there is no value of paired with -coordinate 0, and, subsequently, the graph of the equation has no -intercept.

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Example Question #2 : How To Find X Or Y Intercept

Give the -intercept, if there is one, of the graph of the equation

$y = \frac{3}{\left | 3x-8\right |}$

Possible Answers:

The graph has no -intercept.

$\left ( 0, \frac{3}{8}\right )$

$\left ( 0, \frac{3}{5}\right )$

$\left ( 0, -\frac{3}{8}\right )$

$\left ( 0,- \frac{3}{5}\right )$

Correct answer:

$\left ( 0, \frac{3}{8}\right )$

Explanation:

The -intercept is the point at which the graph crosses the -axis; at this point, the -coordinate is 0, so substitute for in the equation:

$y = \frac{3}{\left | 3x-8\right |}$

$y = \frac{3}{\left | 3 \cdot 0 -8\right |}$

$y = \frac{3}{\left | 0 -8\right |}$

$y = \frac{3}{\left | -8\right |}$

$y = \frac{3}{ 8}$

The -intercept is $\left ( 0, \frac{3}{8}\right )$ .

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HSPT Math : HSPT Mathematics

Study concepts, example questions & explanations for HSPT Math

All HSPT Math Resources

Example Questions

Example Question #1475 : Concepts

Example Question #1476 : Concepts

Example Question #1477 : Concepts

Example Question #1478 : Concepts

Example Question #1 : How To Find The Endpoints Of A Line Segment

Example Question #2 : How To Find The Endpoints Of A Line Segment

Example Question #1 : How To Find Out If Lines Are Perpendicular

Example Question #3 : Use Similar Triangles To Show Equal Slopes: Ccss.Math.Content.8.Ee.B.6

Example Question #2 : X And Y Intercept

Example Question #2 : How To Find X Or Y Intercept

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