Question

Identify the vertical asymptotes of ula1" title="\frac{x^3-16x}{-3x^2+3x+18}" alt="\frac{x^3-16x}{-3x^2+3x+18}" align="absmiddle" class="latex-formula">
Then sketch the graph.

Answer 1

$\sf \frac{ {x}^{3} - 16x}{ { - 3x}^{2} + 3x + 18 }$

To find the vertical asymptotes set denominator as 0

$\sf - 3x ^{2} + 3x + 18 = 0$

Divide both sides of the equation by-3

$\sf \: {x}^{2} - x - 6 = 0$

Write -x as a difference

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

Factor out x from the expression

$\sf \: x \times (x + 2) - 3x - 6 = 0$

Factor out -3 from the expression

$\sf \: x \times (x + 2) - 3(x + 2) = 0$

Factor out x+2 from the expression

$\sf(x + 2)(x - 3) = 0$

When the product of factors equals 0, at least one factor is 0

$\sf \: x + 2 = 0 \\ \sf \: x - 3 = 0$

Solve the equations for x

$\sf \: x = - 2 \\ \sf \: x = 3$

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• Graph is attached!!~

Answer 2

First we need to check where the function gets undefined.

• It's if denominator is 0

So solve it now

$\\ \tt\hookrightarrow -3x^2+3x+18=0$

$\\ \tt\hookrightarrow -3(x^2-x-6)=0$

$\\ \tt\hookrightarrow x^2-x-6=0$

$\\ \tt\hookrightarrow x^2+2x-3x+6=0$

$\\ \tt\hookrightarrow x(x+2)-3(x+2)=0$

$\\ \tt\hookrightarrow (x-3)(x+2)=0$

$\\ \tt\hookrightarrow x=3\;or\:x=-2$