Question

Determine the slope of the tangent line of the following equation at \(x=0\) and \(x=1\): \(f(x) = -2x^2\). A) At \(x=0\), the slope is 0. At \(x=1\), the slope is -2. B) At \(x=0\), the slope is -2. At \(x=1\), the slope is 0. C) At \(x=0\), the slope is 0. At \(x=1\), the slope is 0. D) At \(x=0\), the slope is -2. At \(x=1\), the slope is -2.

Answer 1

Hi,

0 and -4

Explanation:

`f(x)=-2x^2\\\\f'(x)=-4x\\\\if \ x=0\ , \ f'(0)=-4*0=0:\ slope=0\\\\if \x=1:\ ,\ f'(1)=-4*1=-4:\ slope=-4\\`