Graphing inverse variation - GMAT Quantitative Reasoning

Card 0 of 5

Question

Give the -intercept of the graph of the equation $y = \frac{4}{2x+ 5}$ .

Answer

Set in the equation:

$y = \frac{4}{2x+ 5}$

$y = \frac{4}{2 \cdot 0+ 5} = \frac{4}{ 5}$

The -intercept is $\left (0, \frac{4}{5} \right )$ .

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Question

Give the -intercept(s), if any, of the graph of the equation

$y = \frac{4}{2x+ 5}$

Answer

Set in the equation and solve for .

$y = \frac{4}{2x+ 5}$

$\frac{4}{2x+ 5} = 0$

$\frac{4}{2x+ 5} \cdot (2x+5) = 0 \cdot (2x+5)$

This is impossible, so the equation has no solution. Therefore, the graph has no -intercept.

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Question

Give the vertical asymptote of the graph of the equation

$y = \frac{2x-1}{3x+ 5}$

Answer

The vertical asymptote is , where is found by setting the denominator equal to 0 and solving for :

$3x \div 3 = - 5 \div 3$

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

This is the equation of the vertical asymptote.

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Question

Give the -intercept(s), if any, of the graph of the equation

$y = \frac{2x-7}{3x+5}$

Answer

Set in the equation and solve for .

$y = \frac{2x-7}{3x+5}$

$0 = \frac{2x-7}{3x+5}$

$0 = \frac{2x-7}{3x+5} \cdot (3x+5)$

$2x \div 2 = 7 \div 2$

$x = \frac{7}{2}$

The -intercept is $\left ( \frac{7}{2}, 0 \right )$

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Question

Give the horizontal asymptote, if there is one, of the graph of the equation

$y = \frac{2x-7}{3x+5}$

Answer

To find the horizontal asymptote, we can divide both numerator and denominator in the right expression by :

$y = \frac{2x-7}{3x+5}$

$y = \frac{\frac{2x-7}{x}}{\frac{3x+5}{x}}$

$y = \frac{2- \frac{7}{x}}{3+ \frac{5}{x}}$

As approaches positive or negative infinity, $\frac{7}{x}$ and $\frac{5}{x}$ both approach 0. Therefore, approaches $\frac{2 }{3 }$ , making the horizontal asymptote the line of the equation $y = \frac{2 }{3 }$ .

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