Question

Hi tutor,At given point (-1.5,5) please find the slope of a tangent of this function (and show the derivation steps) much appreciated

Answer 1

Solution:

Given the function below

$f\left(x\right)=2x^3+x^2-5x+2$

To find the slope of a tangent, first find the first derivative, as shown below

$\begin{gathered} \frac{d}{dx}\left(2x^3+x^2-5x+2\right) \\ \frac{dy}{dx}=3(2x^{3-1})+2(x^{2-1})-(5x^{1-1})+0 \\ \frac{dy}{dx}=6x^2+2x-5 \end{gathered}$

At a given point (-1.5,5), the slope of a tangent of this function is

$\begin{gathered} \frac{dy}{dx}=6x^2+2x-5 \\ =6\left(-1.5\right)^2+2\left(-1.5\right)-5 \\ =13.5-3-5=13.5-8=5.5 \\ \frac{dy}{dx}=5.5 \end{gathered}$

Hence, the slope of a tangent of the function is 5.5