Respuesta :

genan

Answer:

vertical asymptotes is x = -2 and x = 1

Step-by-step explanation:

For rational functions, the vertical asymptotes are the undefined points, also known as the zeros of the denominator (below the line), of the simplified function.

This means [tex]-4x^{2} - 4x + 8[/tex] is equal to zero than solve

*image below*

Hope it helps :)

Let me know if you have any questions or anymore help lmk, happy to help with any school work!!

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ItzHeartRider

We are give with the function:

[tex]{\quad \qquad \longrightarrow f(x)=\dfrac{x^{2}+x-12}{-4x^{2}-4x+8}}[/tex]

Vertical asymptotes are the points, where the function is not defined, so the denominator must be equal to 0 for making f(x) not defined

[tex]{:\implies \quad \sf -4x^{2}-4x+8=0}[/tex]

Dividing both sides by -4, we have;

[tex]{:\implies \quad \sf x^{2}+x-2=0}[/tex]

Can be further written as:

[tex]{:\implies \quad \sf x^{2}-x+2x-2=0}[/tex]

[tex]{:\implies \quad \sf x(x-1)+2(x-1)=0}[/tex]

[tex]{:\implies \quad \sf (x-1)(x+2)=0}[/tex]

Using zero product rule, Equating both multiplicants to 0, we have:

[tex]{:\implies \quad \boxed{\bf{x=1\quad or\quad -2}}}[/tex]

This is the required answer

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