Question

select the x - intercept(s) and y - intercept of the function, if one exists. f(x)=\frac{3x - 4}{3x^{2}-x - 4} y - int: none x - int: (\frac{4}{3},0) x - int: none y - int: (0,\frac{4}{3}) x - int: (1,0) y - int: (0,1)

Explanation:

Step1: Find the x - intercept

Set $y = f(x)=0$. So we set the numerator equal to 0 since $\frac{a}{b}=0$ when $a = 0$ and $b
eq0$.
$3x - 4=0$
Solve for $x$:
$3x=4$, then $x=\frac{4}{3}$.
We need to check if the denominator is non - zero at $x = \frac{4}{3}$.
Substitute $x=\frac{4}{3}$ into the denominator $3x^{2}-x - 4$:
$3\times(\frac{4}{3})^{2}-\frac{4}{3}-4=3\times\frac{16}{9}-\frac{4}{3}-4=\frac{16}{3}-\frac{4}{3}-4=\frac{16 - 4}{3}-4=4 - 4=0$. So $x=\frac{4}{3}$ is not an x - intercept.
Factor the denominator $3x^{2}-x - 4=(3x - 4)(x + 1)$. The function can be written as $f(x)=\frac{3x - 4}{(3x - 4)(x + 1)}$ ($x
eq\frac{4}{3}$ and $x
eq - 1$). After canceling out the common factor $(3x - 4)$ (for $x
eq\frac{4}{3}$), we have $f(x)=\frac{1}{x + 1}$ ($x
eq\frac{4}{3}$).
Set $y = 0$ for the simplified function $\frac{1}{x+1}=0$. Since the numerator is 1, there is no value of $x$ that makes $y = 0$. So $x - int:none$.

Step2: Find the y - intercept

Set $x = 0$ in the original function $f(x)=\frac{3x - 4}{3x^{2}-x - 4}$.
$f(0)=\frac{3\times0 - 4}{3\times0^{2}-0 - 4}=\frac{-4}{-4}=1$. So the $y - int:(0,1)$.

Answer:

$x - int:none$
$y - int:(0,1)$