Question

what is the value of $\frac{d}{dx}(2x^{2}-6x + 5)$ at $x=-4$? choose 1 answer: a 61 b -14 c -10 d -22

Explanation:

Step1: Differentiate the function

Use the power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$. For $y = 2x^2-6x + 5$, $\frac{dy}{dx}=\frac{d}{dx}(2x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)$. So, $\frac{dy}{dx}=4x-6$.

Step2: Substitute $x=-4$

Substitute $x = - 4$ into $\frac{dy}{dx}$. We get $\frac{dy}{dx}\big|_{x=-4}=4\times(-4)-6$.
Calculate $4\times(-4)-6=-16 - 6=-22$.

Answer:

D. - 22