Question

Help with homework lesson for review to study for test.

Answer 1

We have the expression:

$\frac{y^2-9}{y^2+y-12}$

The restricted values of y are those values that make the denominator 0. This is the same as solving the equation:

$y^2+y-12=0$

This term can be factorized as:

$\begin{gathered} y^2+y-12=(y+4)(y-3) \\ \Rightarrow(y+4)(y-3)=0 \end{gathered}$

This is true for y = -4 and y = 3. So the restricted values for y are:

$y\ne\mleft\lbrace-4,3\mright\rbrace$