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Describe the vertical asymptote(s) and hole(s) for the graph of y = (x-5)(x-2) over (x-2)(x+4).

Sagot :

dln
To find where there are holes, we need to make the function = [tex] \frac{0}{0} [/tex].
[tex] \frac{(x-5)(x-2)}{(x-2)(x+4)} [/tex]
If you plug in [tex]x=2[/tex], you get [tex] \frac{0}{0} [/tex], which means there is a hole in the graph when [tex]x=2[/tex]

To find where there are vertical asymptotes, we need to make just the bottom 0.
Ex. [tex] \frac{1}{0} [/tex]
[tex] \frac{(x-5)(x-2)}{(x-2)(x+4)} [/tex]
If you plug in [tex]x=-4[/tex], you get [tex] \frac{(-9)(-6)}{0} = \frac{54}{0} [/tex], which means there is a vertical asymptote when [tex]x=-4[/tex]
awhitsett08

Answer:

X=-3 ,-1

Step-by-step explanation:

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