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