Please help,
Find the horizontal & vertical asymptotes (if any) for: h(x)=3/(x-6)(2x-2)
What I think the answer is: No horizontal asymptote, and the vertical would be:
6 & 2??
thank you!!
incorrect. it is true that x = 6 is a vertical asymptote, but everything else you said is wrong.
to find vertical asymptotes, set the denominator to zero and solve for the values of x
that is, solve $\displaystyle (x - 6)(2x - 2) = 0$
for the horizontal asymptotes find the infinite limits of the function (you are in precalculus right? do you know what limits are?)
if you don't know about limits, use the rule that:
if the power of x in the numerator is equal to the power of x in the denominator then the horizontal asymptote is the ratio of the coefficients, that is y = (coefficient of the top)/(coefficient of the bottom) is the horizontal asymptote.
if the power of x in the numerator is less than the power of x in the denominator, then the horizontal asymptote is y = 0
if the power of x in the top is greater than the power of x in the bottom, then there is no horizontal asymptote
Hello, crazydaizy78!
Find the horizontal & vertical asymptotes (if any) for: $\displaystyle h(x)\:=\:\frac{3}{(x-6)(2x-2)}$
What I think the answer is: No horizontal asymptote . . . . wrong
and the vertical would be: 6 & 2 . . . . no
For a horizontal asymptote, consider the limit as $\displaystyle x\to\infty$
. . $\displaystyle \lim_{x\to\infty}h(x) \;=\;\lim_{x\to\infty}\,\frac{3}{(x-6)(2x-2)} \;=\;0$
Therefore: .$\displaystyle y \,=\,0$ (x-axis) is a horizontal asymptote.
For vertical asymptotes: what values of $\displaystyle x$ produce a zero in the denominator?
. . $\displaystyle (x-6)(2x-2) \:=\:0\quad\Rightarrow\quad\begin{array}{ccc}x-6 \:=\:0 & \;\Rightarrow\; & \boxed{x \:=\:6} \\
2x - 2 \:=\:0 & \;\Rightarrow\; & \boxed{x \:=\:1} \end{array} $