Chapter 2.1, Problem 55E

To determine

To find: The equation for the parabola in the standard form.

Answer to Problem 55E

The standard form of equation for the parabola is $f\left(x\right)=\frac{-16}{3}{\left(x+\frac{5}{2}\right)}^2$ .

Explanation of Solution

Given:

The vertex of the parabola is $\left(\frac{-5}{2},0\right)$ and it passes through the point $\left(-\frac{7}{2},\frac{-16}{3}\right)$

Calculation:

Consider the quadratic function is,

  $f\left(x\right)=a{\left(x-h\right)}^2+k$

The vertex of the above parabola for the standard form $f\left(x\right)=a{\left(x-h\right)}^2+k$ is,

  $\left(h,k\right)$

Then, the equation of the parabola with the vertex marked is,

  $\begin{array}{c}f\left(x\right)=a{\left(x-\left(\frac{-5}{2}\right)\right)}^2+0\\ =a{\left(x+\frac{5}{2}\right)}^2\end{array}$ ..............(I)

The parabola passes through the point $\left(-\frac{7}{2},\frac{-16}{3}\right)$ and it follows $f\left(-\frac{7}{2}\right)=\frac{-16}{3}$ . So,

  $\begin{array}{l}f\left(-\frac{7}{2}\right)=a{\left(-\frac{7}{2}+\frac{5}{2}\right)}^2\\ \frac{-16}{3}=a{\left(\frac{-2}{2}\right)}^2\\ a=\frac{-16}{3}\end{array}$

Substitute $\frac{-16}{3}$ for $a$ in equation (I).

  $f\left(x\right)=\frac{-16}{3}{\left(x+\frac{5}{2}\right)}^2$

Thus, the standard form of equation for the parabola is $f\left(x\right)=\frac{-16}{3}{\left(x+\frac{5}{2}\right)}^2$ .