Graphing inverse variation - GMAT Quantitative

Card 0 of 40

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

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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 )$ .

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

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

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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.

Give the vertical asymptote of the graph of the equation

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

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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.

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

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

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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 )$

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

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

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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 }$$ .

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

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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 )$ .

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

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

Tap to see back →

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.

Give the vertical asymptote of the graph of the equation

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

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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.

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

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

Tap to see back →

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 )$

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

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

Tap to see back →

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 }$$ .

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

Tap to see back →

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 )$ .

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

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

Tap to see back →

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.

Give the vertical asymptote of the graph of the equation

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

Tap to see back →

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.

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

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

Tap to see back →

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 )$

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

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

Tap to see back →

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 }$$ .

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

Tap to see back →

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 )$ .

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

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

Tap to see back →

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.

Give the vertical asymptote of the graph of the equation

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

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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.

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

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

Tap to see back →

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 )$

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

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

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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 }$$ .