Chapter 2.1, Problem 50E

To determine

To find:The x-intercept(s) of quadratic function using graphing utility, and algebraically to verify the answer.

Answer to Problem 50E

The x- intercepts of the graph of quadratic function $y=\frac{7}{10}\left(x^2+12x-45\right)$ are 3 and $-15$ .

Explanation of Solution

Given data: $y$

The given quadratic function is $y=\frac{7}{10}\left(x^2+12x-45\right)$ .

Concept used:

The x-intercept (s) of the graph of a function $y=f\left(x\right)$ are the values of variable “x” for which $y=f\left(x\right)=0$ . If the graph of function $y=f\left(x\right)$ intersect x-axis at point $\left(a\text{,0}\right)$ then the x- intercept of the function is “a”.

Calculations:

The graph of the function is shown in Figure-1.

We see that the graph of the quadratic function $y=\frac{7}{10}\left(x^2+12x-45\right)$ intersects the x-axisat points $\text{A}\left(\text{3,0}\right)$ and $\text{B}\left(\text{-15,0}\right)$ .

Hence, the x- intercepts of the graphof quadratic function $y=\frac{7}{10}\left(x^2+12x-45\right)$ are 3 and $-1$ 5.

Algebraically, the x-intercept (s) of the graph $y=\frac{7}{10}\left(x^2+12x-45\right)$ are obtained by solving equation $y=0$

  $\Rightarrow \frac{7}{10}\left(x^2+12x-45\right)=0$

  $\Rightarrow x^2+12x-45=0$

  $\Rightarrow x^2+15x-3x-45=0$

  $\Rightarrow x\left(x+15\right)x-3\left(x+15\right)=0$

  $\Rightarrow \left(x-3\right)\left(x+15\right)=0$

  $\Rightarrow \left(x-3\right)=0,\left(x+15\right)=0$

  $\Rightarrow x=3,x=-15$ .

Thus, the x- intercepts of the graph of quadratic function $y=\frac{7}{10}\left(x^2+12x-45\right)$ are 3and $-15$

Conclusion:

The x- intercepts of the graph of quadratic function $y=\frac{7}{10}\left(x^2+12x-45\right)$ are 3 and $-15$ .