Question

For the following exercise, use the graph to write an equation for the function. (Please write in expanded form and factored form)

Answer 1

We are asked to determine a function that has two vertical asymptotes at:

$\begin{gathered} x=-3 \\ x=4 \end{gathered}$

And an x-intercept at:

$x=3$

And a y-intercept at:

$y=-2$

This means that the function must have the following form:

$f(x)=\frac{k(x-x_0)}{(x-a)(x-b)}$

Where:

$\begin{gathered} a,b=\text{ vertical asymptotes} \\ x_0=\text{ x-intercept} \\ k=\text{ constant} \end{gathered}$

Now, we substitute the known values:

$f(x)=\frac{k(x-3)}{(x+3)(x-4)}$

Now, we determine the value of "k" using the y-intercept, since this means that when "x = 0", then "y = -2". Substituting we get:

$-2=\frac{k(0-3)}{(0+3)(0-4)}$

Solving the operations:

$-2=\frac{-3k}{(3)(-4)}$

Simplifying:

$-2=\frac{-k}{-4}$

Now, we multiply both sides by -4:

$8=-k$

Now, we multiply both sides by -1:

$-8=k$

Substituting in the function we get:

$f(x)=\frac{-8(x-3)}{(x+3)(x-4)}$

And thus we get the function we were looking for in the factored form.

Now, to determine the expanded form we use the distributive property on the denominator, we get:

$f(x)=\frac{-8(x-3)}{x^2-4x+3x-12}$

Adding like terms:

$f(x)=\frac{-8(x-3)}{x^2-x-12}$

Now, we apply the distributive property on the numerator:

$f(x)=\frac{-8x+24}{x^2-x-12}$

And thus we get the expanded form.