Q. 320 - Intermediate Algebra

Question

In the following exercises, for each function, find:

a. the zeros of the function

b. the $x$- intercepts of the function

c. the $y$- intercept of the function

$f\left(x\right)=12x^2-11x+2$

Step-by-Step Solution

Answer

a. The zeros of the function are $\frac{2}{3}, \frac{1}{4}$.

b. The $x$- intercepts are $(\frac{2}{3}, 0)$ and $(\frac{1}{4}, 0)$.

c. The $y$- intercept is $(0, 2)$.

Part a. Finding zeros of the function

To find the zeros of the function, we need to find when the function value is $0$.

We substitute $0$ for $f\left(x\right)$ in the given equation.

$$\begin{gathered} f\left(x\right)=12x^2-11x+2 \\ 0=12x^2-11x+2 \\ 12x^2-11x+2=0 \\ 12x^2-8x-3x+2=0 \\ 4x(3x-2)-1(3x-2)=0 \\ (3x-2)(4x-1)=0 \end{gathered}$$

Using the Zero Product Property,

$3x-2=0$ or

$4x-1=0$.

When $3x-2=0$,

$x=\frac{2}{3}$.

When $4x-1=0$,

$x=\frac{1}{4}$.

These are the zeros of the function.

Part b. The x - intercepts of the graph of the function

An $x$- intercept occurs when $y=0$.

We know that,

$f\left(\frac{2}{3}\right)=0$ 

$f\left(\frac{1}{4}\right)=0$.

Thus, the points $(\frac{2}{3}, 0)$ and $(\frac{1}{4}, 0)$ lie on the graph of the function.

Hence, these points are the $x$- intercepts of the function. 

Part c. The y - intercepts of the graph of the function

A $y$- intercept of the graph of the function occurs when $x=0$.

To find it, we find $f\left(0\right)$.

$$\begin{gathered} f\left(x\right)=12x^2-11x+2 \\ f\left(0\right)=12{\left(0\right)}^2-11\left(0\right)+2 \\ f\left(0\right)=2 \end{gathered}$$

Since $f\left(0\right)=2$, the point $(0, 2)$ lies on the graph and is the $y$- intercept of the function.