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

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} (d)/(dx)\left(2x^3+x^2-5x+2\right) \\ (dy)/(dx)=3(2x^(3-1))+2(x^(2-1))-(5x^(1-1))+0 \\ (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} (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 \\ (dy)/(dx)=5.5 \end{gathered}`

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