39)
Its always best to factorise where you can, in most questions it tends to be a necessary step and even if its not, it usually makes things easier;
In this case, you can factorise the denominator:
[tex]h(x) = \frac{x \ - \ 1}{ x^{2} \ - \ x \ - \ 12} \\\\ = \frac{x \ - \ 1}{(x \ - \ 4)(x \ + \ 3)}[/tex]
To get y-intercept/s, set x = 0 and solve:
[tex]h(0) = \frac{(0) \ - \ 1}{((0) \ - \ 4)((0) \ + \ 3)} \\\\ = \frac{-1}{-12} \\\\ = \frac{1}{12}[/tex]
y-intercept: (0, 1/12)
To get x-intercept/s, set y = 0 and solve:
[tex]\frac{x \ - \ 1}{(x \ - \ 4)(x \ + \ 3)} = 0 \\\\ x \ - \ 1 = 0 \\\\ x = 1[/tex]
x-intercept: (1, 0)
To find the asymptotes, it is necessary for us to have factorised the denominator so its a good thing that I did so at the beginning;
To get the vertical asymptote/s, set the denominator equal to 0, then just rearrange to get x:
[tex](x \ - \ 4)(x \ + \ 3) = 0 \\ x \ - \ 4 = 0 \\ x = 4 \\ and \\ x \ + \ 3 = 0 \\ x = -3[/tex]
vertical asymptotes: x = 4 and x = -3
There are no horizontal asymptotes.