Question

let $f(x) = \frac{2x^2 - 1x - 3}{2x^2 - 5x - 12}$
this function has:

1. a $y$ intercept at the point
2. $x$ intercepts at the point(s)
3. vertical asymptotes at $x = $
4. horizontal asymptote at $y = $

Explanation:

1) y - intercept

Step 1: Recall y - intercept definition

The y - intercept of a function \(y = f(x)\) is found by setting \(x = 0\) and finding the corresponding \(y\) - value.

Step 2: Substitute \(x = 0\) into \(f(x)\)

$$
LATEXBLOCK0
$$

Step 1: Recall x - intercept definition

The x - intercepts of a function \(y = f(x)\) are found by setting \(y=f(x) = 0\) (i.e., numerator \(= 0\) and denominator \(
eq0\)).

Step 2: Solve the numerator \(2x^{2}-x - 3=0\)

Factor the quadratic equation: \(2x^{2}-x - 3=(2x - 3)(x + 1)=0\)
Set each factor equal to zero:

• \(2x-3 = 0\Rightarrow x=\frac{3}{2}\)
• \(x + 1=0\Rightarrow x=-1\)

Now, we need to check if these values make the denominator zero.
For \(x=\frac{3}{2}\), denominator \(2(\frac{3}{2})^{2}-5(\frac{3}{2})-12=2\times\frac{9}{4}-\frac{15}{2}-12=\frac{9}{2}-\frac{15}{2}-12=- 3 - 12=-15
eq0\)
For \(x=-1\), denominator \(2(-1)^{2}-5(-1)-12=2 + 5-12=-5
eq0\)

Step 1: Recall vertical asymptote definition

Vertical asymptotes occur where the denominator is zero and the numerator is not zero.

Step 2: Solve the denominator \(2x^{2}-5x - 12 = 0\)

Factor the quadratic equation: \(2x^{2}-5x - 12=(2x + 3)(x - 4)=0\)
Set each factor equal to zero:

• \(2x+3=0\Rightarrow x=-\frac{3}{2}\)
• \(x - 4=0\Rightarrow x = 4\)

Now, check the numerator at these values:
For \(x=-\frac{3}{2}\), numerator \(2(-\frac{3}{2})^{2}-(-\frac{3}{2})-3=2\times\frac{9}{4}+\frac{3}{2}-3=\frac{9}{2}+\frac{3}{2}-3=6 - 3 = 3
eq0\)
For \(x = 4\), numerator \(2(4)^{2}-4 - 3=32-4 - 3=25
eq0\)

Answer:

\((0,\frac{1}{4})\)

2) x - intercepts