Question

Find the slope of the tangent to the curve $y=4x^2-3x-8$ at point $(-3,4)$.

• A. $27$
• B. $-27$
• C. $-29$
• D. $29$

Answer 1

The correct option is B $-27$

The slope of the tangent to the curve $y=4x^2-3x-8$ is given by $\frac{dy}{dx}$
So, we have slope as
$\frac{dy}{dx}=\frac{d(4x^2-3x-8)}{dx}=8x-3$
Thus, slope of the tangent at point $(-3,4)$ is given as
$\frac{dy}{dx}{|}_{(-3,4)}=(8\times (-3)-3)=-27$