Chapter 3, Problem 15CLT

To determine

To write: standard form of the quadratic function whose graph is a parabola with vertex $\left(-8,5\right)$ and that passes through the point $\left(-4,-7\right)$ .

Answer to Problem 15CLT

  $f\left(x\right)=-\frac{3}{4}{\left(x+8\right)}^2+5$

Explanation of Solution

Concept Used:

The standard form of the quadratic function f whose graph is a parabola whose axis is the vertical line $x=h$ and whose vertex is the point $\left(h,k\right)$ is:

  $f\left(x\right)=a{\left(x-h\right)}^2+k$ , $a\ne 0$

It is given that the vertex of parabola is at $\left(-8,5\right)$ , so the equation has the form:

  $f\left(x\right)=a{\left(x-\left(-8\right)\right)}^2+5$

Since, parabola passes through $\left(-4,-7\right)$ , it implies $f\left(-4\right)=-7$ . So,

  $\begin{array}{c}f\left(-4\right)=a{\left(-4-\left(-8\right)\right)}^2+5\\ -7=a{\left(-4+8\right)}^2+5\\ -7-5=a{\left(4\right)}^2\end{array}$

  $\begin{array}{c}-12=16a\\ -\frac{12}{16}=a\\ -\frac{3}{4}=a\end{array}$

Thus, the function in standard form is written as:

  $f\left(x\right)=-\frac{3}{4}{\left(x+8\right)}^2+5$