Q. 318 - Intermediate Algebra

Question

In the following exercises, for each function, find:

a. the zeros of the function

b. the $x$- intercepts of the graph of the function.

c. the $y$- intercept of the graph of the function.

$f\left(x\right)=25x^2-49$

Step-by-Step Solution

Answer

a. The zeros are $\frac{7}{5}, -\frac{7}{5}$.

b. The $x$- intercepts are $(\frac{7}{5}, 0)$ and $(\frac{-7}{5}, 0)$.

c. The $y$- intercept is $(0, -49)$.

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.

$$\begin{gathered} f\left(x\right)=25x^2-49 \\ 0=25x^2-49 \\ {x}^2=\frac{49}{25} \\ x=\frac{7}{5} \end{gathered}$$

or $x=-\frac{7}{5}$

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,

$f\left(\frac{7}{5}\right)=0$ and

$f(-\frac{7}{5})=0$.

Thus, the points $(\frac{7}{5}, 0)$ and $(-\frac{7}{5}, 0)$ lie on the graph of the function.

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

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

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

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

$$\begin{gathered} f\left(x\right)= 25x^2-49 \\ f\left(0\right)=25{\left(0\right)}^2-49 \\ f\left(0\right)=-49 \end{gathered}$$

Since, $f\left(0\right)=-49$, the point $(0, -49)$ lies on the graph and is the $y$- intercept of the function.