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SSAT Upper Level Math : How to graph inverse variation

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Example Questions

Example Question #1 : How To Graph Inverse Variation

Give the equation of the vertical asymptote of the graph of the equation $y = \frac{7}{9x+4}$ .

Possible Answers:

$x= \frac{7}{9}$

$x= - \frac{7}{4}$

x = 9

$x= -\frac{4}{9}$

x=-4

Correct answer:

$x= -\frac{4}{9}$

Explanation:

The vertical asymptote of an inverse variation function is the vertical line of the equation x = a , where is the value for which the expression is not defined. To find , set the denominator to and solve for :

9x+4 = 0

9x= -4

$x= -\frac{4}{9}$ is the equation of the asymptote.

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Example Question #2 : How To Graph Inverse Variation

Give the -intercept of the graph of the equation $y = \frac{5}{2x-7}$ .

Possible Answers:

$\left ( \frac{7}{2}, 0 \right )$

(7,0)

The graph has no -intercept.

(5,0)

$\left ( \frac{5}{2}, 0 \right )$

Correct answer:

The graph has no -intercept.

Explanation:

The -intercept of the graph of an equation is the point at which it intersects the -axis. Its -coordinate is 0, so set y = 0 and solve for :

$y = \frac{5}{2x-7}$

$0= \frac{5}{2x-7}$

$0\cdot (2x-7) = \frac{5}{2x-7} \cdot (2x-7)$

0 = 5

This is identically false, so the graph has no -intercept.

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Example Question #3 : How To Graph Inverse Variation

Give the -intercept of the graph of the equation $y = \frac{3}{5x-8}$ .

Possible Answers:

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

The graph has no -intercept.

$\left ( 0, \frac{8}{5}\right )$

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

(0,8)

Correct answer:

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

Explanation:

The -intercept of the graph of an equation is the point at which it intersects the -axis. Its -coordinate is 0, so set x = 0 and solve for :

$y = \frac{3}{5x-8}$

$y = \frac{3}{5 (0)-8}$

$y = \frac{3}{0-8}$

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

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

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Example Question #4 : How To Graph Inverse Variation

Give the slope of the line that passes through the - and -intercepts of the graph of the equation $y = \frac{6}{8x-2}$ .

Possible Answers:

The line cannot exist as described.

$-\frac{1}{3}$

$\frac{3}{4}$

$\frac{4}{3}$

Correct answer:

The line cannot exist as described.

Explanation:

The graph of $y = \frac{6}{8x-2}$ does not have an -intercept. If it did, then it would be the point on the graph with -coordinate 0. If we were to make this substitution, the equation would be

$0= \frac{6}{8x-2}$

and

$0 \cdot (8x-2)= \frac{6}{8x-2} \cdot (8x-2)$

0=6

This is identically false, so the graph has no -intercept. Therefore, the line cannot exist as described.

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Example Question #5 : How To Graph Inverse Variation

Give the -coordinate of a point with a positive -coordinate at which the graphs of the equations $y = \frac{4}{x}$ and x+3y = 4 intersect.

Possible Answers:

$\frac{2}{3}$

$\frac{4}{3}$

The graphs of the equations do not intersect.

Correct answer:

The graphs of the equations do not intersect.

Explanation:

Substitute $\frac{4}{x}$ for in the second equation:

$x+3\cdot \frac{4}{x} = 4$

$x+ \frac{12}{x} = 4$

$x+ \frac{12}{x}- 4 = 4 - 4$

$x- 4 + \frac{12}{x}= 0$

$\left (x- 4 + \frac{12}{x} \right )\cdot x = 0 \cdot x$

$x^{2}- 4 x + 12 = 0$

The discriminant of this quadratic expression is $b^{2} - 4ac$ , where a=1, b= -4, c=12 ; this is

$b^{2} - 4ac = (-4)^{2}-4 \cdot 1 \cdot 12 = 16 -48 = -32$ .

The discriminant being negative, there are no real solutions to this quadratic equation. Consequently, there are no points of intersection of the graphs of the two equations on the coordinate plane.

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Example Question #6 : How To Graph Inverse Variation

The graphs of the equations $y = \frac{3}{x}$ and 4x-2y = -10 intersect in two points; one has a positive -coordinate. and one has a negative -coordinate. Give the -coordinate of the point of intersection that has a positive -coordinate.

Possible Answers:

Correct answer:

Explanation:

Substitute $\frac{3}{x}$ for in the second equation:

4x-2y = -10

$4x-2 \left ( \frac{3}{x} \right )= -10$

$4x- \frac{6}{x} = -10$

$4x- \frac{6}{x} + 10 = -10+ 10$

$4x- \frac{6}{x} + 10 = 0$

$\left (4x- \frac{6}{x} + 10 \right ) \cdot \frac{x}{2} = 0 \cdot \frac{x}{2}$

$2x^{2}-3+5x = 0$

$2x^{2}+5x -3= 0$

This quadratic equation can be solved using the method; the integers with product $2 \cdot (-3) = -6$ and sum 5 are and 6, so continue as follows:

$2x^{2}-x+6x -3= 0$

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

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

Either x+3 = 0 , in which case x = -3 , or

2x-1 = 0 , in which case

2x = 1

x = 0.5

The desired -coordinate is paired with the positive -coordinate, so we substitute 0.5 for in the first equation:

$y = \frac{3}{x} = \frac{3}{0.5} = 6$

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Example Question #5 : How To Graph Inverse Variation

Give the -coordinate of the point at which the graphs of the equations $y = \frac{2}{x}$ and $y = -\frac{3}{x+1}$ intersect.

Possible Answers:

The graphs of the equations do not intersect.

Correct answer:

Explanation:

Using the substitution method, set the values of equal to each other.

$\frac{2}{x} = -\frac{3}{x+1}$

Multiply both sides by $\frac{x(x+1)}{1}$ :

$\frac{2}{x} \cdot \frac{x(x+1)}{1}= -\frac{3}{x+1}\cdot \frac{x(x+1)}{1}$

2(x+1)= -3x

2x+2= -3x

2x+2 + 3x -2 = -3x + 3x -2

5x = -2

$5x \div 5 = -2 \div 5$

x = -0.4

Substitute in either equation:

$y = \frac{2}{x} = \frac{2}{-0.4} = -5$

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Example Question #6 : How To Graph Inverse Variation

A line with slope 4 shares its -intercept with that of the graph of the equation $y = \frac{3}{2x-7}$ . Which of the following is the equation of that line?

Possible Answers:

This line does not exist, since the graph of $y = \frac{3}{2x-7}$ has no -intercept.

8x-2y =- 7

28x-7y=2

8x-2y=-3

28x-7y=3

Correct answer:

28x-7y=3

Explanation:

The -intercept of the graph of $y = \frac{3}{2x-7}$ —the point at which it crosses the -axis—is the point at which x = 0 , so substitute accordingly and solve for :

$y = \frac{3}{2 \cdot 0-7}= \frac{3}{ 0-7}=- \frac{3}{ 7}$

The -intercept of this graph, and that of the line, is $\left (0, - \frac{3}{ 7} \right )$ . Since the slope is 4, the slope-intercept form of the equation of the line is

$y = 4x-\frac{3}{7}$

To put it in standard form:

$y-y + \frac{3}{7} = 4x-\frac{3}{7} -y + \frac{3}{7}$

$\frac{3}{7} = 4x -y$

$\frac{3}{7} \cdot 7 = (4x -y) \cdot 7$

28x-7y=3

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Example Question #7 : How To Graph Inverse Variation

$\bigtriangleup ABC \cong \bigtriangleup DEF$ , where $\angle B$ is a right angle, $m \angle C = 30 ^{\circ }$ , and AC = 30 . Which of the following cannot be true?

Possible Answers:

All of the statements given in the other choices are true.

$\angle F$ is a right angle

$m \angle D = 60^{\circ }$

$\bigtriangleup DEF$ has perimeter $45+15 \sqrt{3}$

DE= 15

Correct answer:

$\angle F$ is a right angle

Explanation:

$\angle B$ is a right angle and $m \angle C = 30 ^{\circ }$ , so

$m \angle A = 90 ^{\circ } - m \angle C = 90 ^{\circ } - 30 ^{\circ }= 60 ^{\circ }$ ,

making $\bigtriangleup ABC$ a 30-60-90 triangle. By the 30-60-90 Triangle Theorem, the length of the short leg $\overline{AB}$ is half that of hypotenuse $\overline{AC}$ :

$AB= \frac{1}{2}\cdot AC = \frac{1}{2}\cdot 30 = 15$

and the length of long leg $\overline{BC}$ is $\sqrt {3}$ times that of $\overline{AB}$ :

$BC = AB \cdot \sqrt{3}= 15 \cdot \sqrt{3}$

Corresponding sides of congruent triangles are congruent, so DE = AB ; since AB=15 , it follows that DE = 15 .

Also, DE = AB = 15 , $EF= BC = 15 \sqrt{3}$ , and DF = AC = 30 , so the perimeter of $\bigtriangleup DEF$ is the sum of these, or

$DE+EF+DF = 15+15 \sqrt{3}+30=45+15 \sqrt{3}$ .

Corresponding angles are congruent, so $m \angle D = m \angle A$ and $m \angle F = m \angle C$ . By substitution, $m \angle D = 60 ^{\circ }$ and $m \angle F = 30 ^{\circ }$ .

The false statement among the choices is that $\angle F$ is a right angle.

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SSAT Upper Level Math : How to graph inverse variation

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #1 : How To Graph Inverse Variation

Example Question #2 : How To Graph Inverse Variation

Example Question #3 : How To Graph Inverse Variation

Example Question #4 : How To Graph Inverse Variation

Example Question #5 : How To Graph Inverse Variation

Example Question #6 : How To Graph Inverse Variation

Example Question #5 : How To Graph Inverse Variation

Example Question #6 : How To Graph Inverse Variation

Example Question #7 : How To Graph Inverse Variation

All SSAT Upper Level Math Resources

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