Chapter 2.1, Problem 56E

To determine

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

Answer to Problem 56E

The standard form of equation for the parabola is $f\left(x\right)=-450{\left(x-6\right)}^2+6$ .

Explanation of Solution

Given:

The vertex of the parabola is $\left(6,6\right)$ and it passes through the point $\left(\frac{61}{10},\frac{3}{2}\right)$

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,

  $f\left(x\right)=a{\left(x-6\right)}^2+6$ ..............(I)

The parabola passes through the point $\left(\frac{61}{10},\frac{3}{2}\right)$ and it follows $f\left(\frac{61}{10}\right)=\frac{3}{2}$ . So,

  $\begin{array}{l}f\left(\frac{61}{10}\right)=a{\left(\frac{61}{10}-6\right)}^2+6\\ \frac{3}{2}-6=\frac{a}{100}\\ a=100\left(\frac{-9}{2}\right)\\ a=-450\end{array}$

Substitute $-450$ for $a$ in equation (I).

  $f\left(x\right)=-450{\left(x-6\right)}^2+6$

Thus, the standard form of equation for the parabola is $f\left(x\right)=-450{\left(x-6\right)}^2+6$ .