# interpreting a function

• May 18th 2010, 07:43 PM
foreverbrokenpromises
interpreting a function
I do not know how to do this question
Consider the function $\displaystyle f(x) = [(2x-a)(x-b)]/[5(x-b)(x-c)]$ . Determine where the following features of its graph are located in terms of a, b, and/or c. Explain your reasoning for each situation.

a. the x-intercepts

b. the y-intercepts

c. any vertical asymptotes

d. any horizontal asymptotes

e. any “holes” in the graph

How do I find a b c d and e? does graphing it out make it easier? or are there formulas to use as well?
• May 18th 2010, 08:12 PM
dwsmith
Quote:

Originally Posted by foreverbrokenpromises
I do not know how to do this question
Consider the function $\displaystyle f(x) = [(2x-a)(x-b)]/[5(x-b)(x-c)]$ . Determine where the following features of its graph are located in terms of a, b, and/or c. Explain your reasoning for each situation.

a. the x-intercepts

b. the y-intercepts

c. any vertical asymptotes

d. any horizontal asymptotes

e. any “holes” in the graph

How do I find a b c d and e? does graphing it out make it easier? or are there formulas to use as well?

a. For x intercepts, we need to find points in the form (x,0) so lets do that.
$\displaystyle f(x)=\frac{(2x-a)(x-b)}{5(x-b)(x-c)}=0\rightarrow (2x-a)(x-b)=2x^2-x(2b+a)+ab$$\displaystyle =(x-b)(2x-a)=0$
You can finish it.

b. For y intercepts, we need to find points in the form (0,y).
$\displaystyle f(0)=\frac{(2*0-a)(0-b)}{5(0-b)(0-c)}=\frac{ab}{5bc}$

c. Vertical asymptotes will happen the fraction is undefined $\displaystyle \frac{x}{0}$
$\displaystyle 5(x-b)(x-c)=0\rightarrow x=b,c$

d. For this, we need to run the limits.
$\displaystyle \lim_{x\to\infty}\frac{(2x-a)(x-b)}{5(x-b)(x-c)}\rightarrow \frac{2x^2}{5x^2}=\frac{2}{5}$

e. This is actually giving to you since we are giving the fractions factored. This occurs when $\displaystyle x=b$