Chapter 7.7, Problem 35WE

To determine

Tofind: thequadratic equation.

Answer to Problem 35WE

The quadratic equation is $f\left(x\right)=\frac{2}{5}\left(x^2-6x+5\right)$ .

Explanation of Solution

Given information:

y- intercept $2$ ; x -intercept $1\text{ and 5}$ .

Calculation:

to find the quadratic equation, first use the formula of quadratic formula with roots,

$f\left(x\right)=a\left[x^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)\right]$

Now, substitute and simplify,

$\begin{array}{l}f\left(x\right)=a\left[x^2-\left(1+5\right)x+\left(\left(1\right)\left(5\right)\right)\right]\text{ [substitute]}\\ f\left(x\right)=a\left(x^2-6x+5\right)\text{ [simplify]}\end{array}$

Since, y- intercept $2$ so when $x=0$ then $y=2$ so substitute coordinates in the equation,

$\begin{array}{l}f\left(x\right)=a\left(x^2-6x+5\right)\\ 2=a\left({\left(0\right)}^2-6\left(0\right)+5\right)\text{ [substitute]}\\ 2=a\left(5\right)\text{ [add and subtract]}\\ a=\frac{2}{5}\text{ [simplify]}\end{array}$

Thus, the equation is $f\left(x\right)=\frac{2}{5}\left(x^2-6x+5\right)$ .