Question

what is the slope of the secant line that intersects the graph of $h(x)=16 - x^{2}$ at $x = 1$ and $x = 4$?

Explanation:

Step1: Find function values at given x - values

First, find \(h(1)\) and \(h(4)\).
For \(x = 1\), \(h(1)=16 - 1^{2}=16 - 1 = 15\).
For \(x = 4\), \(h(4)=16 - 4^{2}=16 - 16 = 0\).

Step2: Use slope formula for secant line

The slope \(m\) of the secant line passing through \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Here, \(x_1 = 1,y_1=h(1)=15,x_2 = 4,y_2=h(4)=0\).
So \(m=\frac{0 - 15}{4 - 1}=\frac{- 15}{3}=-5\).

Answer:

\(-5\)