How to graph inverse variation - SSAT Upper Level Quantitative (Math)

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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

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

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

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Question

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

Answer

The -intercept of the graph of an equation is the point at which it intersects the -axis. Its -coordinate is 0, so set 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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Question

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

Answer

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

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

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

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Question

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

Answer

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

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

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

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

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

Substitute in either equation:

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

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Question

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?

Answer

The -intercept of the graph of $y = \frac{3}{2x-7}$ —the point at which it crosses the -axis—is the point at which , 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:

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

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

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

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Question

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

Answer

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 ; 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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Question

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

Answer

$\angle B$ 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 ; since , it follows that .

Also, , $EF= BC = 15 \sqrt{3}$ , and , 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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Question

The graphs of the equations $y = \frac{3}{x}$ and 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.

Answer

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

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

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

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

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

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

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

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

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:

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

Either , in which case , or

, in which case

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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