Chapter 3.4, Problem 5QR

To determine

To find: The vertex of the function by analysing the equation algebraically.

Answer to Problem 5QR

The vertex of the parabola is $\left(5,144\right)$.

Explanation of Solution

Given Information: The equation of the quadratic function is $f\left(x\right)=-16x^2+160x-256$

Calculation:

Consider the equation $f\left(x\right)=-16x^2+160x-256$.

The x -coordinate of the vertex of the parabola of the form $y=a{x}^2+bx+c$ is given by $x=-\frac{b}{2a}$ and y- coordinate is $f\left(x\right)$.

The x- coordinate of the parabola $f\left(x\right)=-16x^2+160x-256$ is,

  $\begin{array}{c}x=-\frac{b}{2a}\\ =-\frac{160}{2\left(-16\right)}\left(a=-16,b=160\right)\\ =5\end{array}$

The x- coordinate represents the value that maximizes the function.

The y -coordinate is,

  $\begin{array}{c}f\left(x\right)=-16x^2+160x-256\\ f\left(5\right)=-16{\left(5\right)}^2+160\left(5\right)-256\\ =-400+800-256\\ =144\end{array}$

Therefore, the vertex of the parabola is $\left(5,144\right)$.

Conclusion:

The vertex of the parabola is $\left(5,144\right)$.