Chapter 8, Problem 13CT

To determine

Tofind:The quadratic function in standard form whose graph satisfies the given conditions.

Answer to Problem 13CT

The quadratic function passes through points $\left(4,0\right)$ and $\left(1,9\right)$ is $f\left(x\right)={\left(x-4\right)}^2$ .

Explanation of Solution

Given information:

The quadratic function passes through points $\left(4,0\right)$ and $\left(1,9\right)$ .

Formula used:

A quadratic function $f\left(x\right)$ having $x=\alpha$ and $x=\beta$ is given by $f\left(x\right)=a\left(x-\alpha \right)\left(x-\beta \right)$ .

Calculations:

Let $f\left(x\right)$ be the quadratic function that passes through points $\left(4,0\right)$ and $\left(1,9\right)$ .

Clearly, point $\left(4,0\right)$ is the x -intercept of $f\left(x\right)$ that implies $f\left(4\right)=0$ . This shows that $x=4$ is zero of $f\left(x\right)$ . Therefore, the function may be written as

  $f\left(x\right)=a\left(x-4\right)\left(x-\beta \right)$.... (1)

The function (1) passes through point $\left(1,9\right)$ that implies $f\left(1\right)=9$ .

Now,   $f\left(1\right)=a\left(1-4\right)\left(1-\beta \right)$

  $\Rightarrow 9=-3a\left(1-\beta \right)$

  $\Rightarrow a\left(1-\beta \right)=-3$.... (2)

Since there are given only two conditions, so for simplicity we take $a=1$ . Then from result (3)

  $\left(1-\beta \right)=-3$

  $\Rightarrow \beta =4$

Substituting $a=1$ and $\beta =4$ in result (1)

  $f\left(x\right)=\left(x-4\right)\left(x-4\right)$

Thus, the quadratic function passes through points $\left(4,0\right)$ And $\left(1,9\right)$ is

  $f\left(x\right)={\left(x-4\right)}^2$ .

Conclusion:

The quadratic function passes through points $\left(4,0\right)$ And $\left(1,9\right)$ Is $f\left(x\right)={\left(x-4\right)}^2$ .