Given f(x) = -2x^2 + 4, determine the slope of the secant line over the interval [-1, 2].
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Answer: -2
Step-by-step explanation:
We know that the slope of a secant line over a interval [a,b] is given by :-
[tex]m=\dfrac{f(b)-f(a)}{b-a}[/tex]
Given f(x) =[tex]-2x^2 + 4[/tex]
Then, the slope of the secant line over the interval [-1, 2] is given by :-
[tex]m=\dfrac{f(2)-f(-1)}{2-(-1)}\\\\=\dfrac{(-2(2)^2+4)-(-2(-1)^2+4)}{2+1}\\\\=\dfrac{(-2(4)+4)-(-2(1)+4)}{3}\\\\=\dfrac{(-8+4)-(-2+4)}{3}\\\\=\dfrac{-4-2}{3}\\\\=\dfrac{-6}{3}\\\\=-1[/tex]
Hence, the slope of the secant line over the interval [-1, 2] is -2.