What is the vertical stretch factor of y=2x^2-6x-8?

idk man good luck man

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation:

vertical asymptote .

horizontal asymptote .

step-by-step explanation:

given function is .

vertical asymptote is given by setting

denominator =0

when degree of numerator and denominator are same then divide first term of numerator by first term of denominator to get horizontal asymptote.

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