given f(x)=(1x-7)(x+2)/(-7x-2)(2x-7)
what is the vertical asymptotes?
what is the horizontal asymptotes?
vertical:
-7x-2 = 0
-7x = 2
x = -2/7
2x-7 = 0
2x = 7
x = 7/2
Horizontal: the highest power of x on the top and on the bottom are the same (x^2), so it would be the coefficient on the top over the coefficient on the bottom.
If you foiled the top you would get x^2... and on the bottom -14x^2...
so -1/14 is the horizontal asymptote.
f(x)=(x^2-5x-14)/(-7x-2)(2x-7)
Vertical asymptote means that for a determined x value, y goes to infinite. So the denominator must be null (but not so for the numberator)
(-7x-2)(2x-7)=0
x=-2/7
x=7/2
for x that goes to -2/7 f(x) goes to -infinite
for x that goes to 7/2 f(x) goes to -infinite
We can find horizontal asyntotes by making x go to + or - infinite: so we put x^2 in evidence
f(x)=x^2(1-5/x-14/x^2)/x^2(-14+45/x+14/x^2)=-1/14
this is the horizontal asyntote
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Verified answer
vertical:
-7x-2 = 0
-7x = 2
x = -2/7
2x-7 = 0
2x = 7
x = 7/2
Horizontal: the highest power of x on the top and on the bottom are the same (x^2), so it would be the coefficient on the top over the coefficient on the bottom.
If you foiled the top you would get x^2... and on the bottom -14x^2...
so -1/14 is the horizontal asymptote.
f(x)=(x^2-5x-14)/(-7x-2)(2x-7)
Vertical asymptote means that for a determined x value, y goes to infinite. So the denominator must be null (but not so for the numberator)
(-7x-2)(2x-7)=0
x=-2/7
x=7/2
for x that goes to -2/7 f(x) goes to -infinite
for x that goes to 7/2 f(x) goes to -infinite
We can find horizontal asyntotes by making x go to + or - infinite: so we put x^2 in evidence
f(x)=x^2(1-5/x-14/x^2)/x^2(-14+45/x+14/x^2)=-1/14
this is the horizontal asyntote