Question

For the function $f\left(x\right)=4x^2-11x-3$

(a) Find the derivative of f at $x=2$.

(b) Find the equation of the tangent line to the graph of $f$at the point$(2,-9)$.

(c) Graph $f$and the tangent line.

Short answer

The derivative of the $f\left(2\right)=5$and the equation at $(2,-9)=y=5x-19$

Step-by-step solution

Part(a) Step 1, Given Information

We are given a function$f\left(x\right)=4x^2-11x-3$

We need to find the derivative of the function at$x=2$$x=2.$

Part(a) Step 2. Explanation

We know,

$f^{\text{'}}\left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$

Also,

$f\left(2\right)=4(2{)}^2-11(2)-3=-9$

Then, using the formula,

$$\begin{gathered} f^{\text{'}}\left(2\right)=\underset{x\to 2}{\lim }\frac{4x^2-11x-3-(-9)}{x-2} \\ =\underset{x\to 2}{\lim }\frac{4x^2-11x+6}{x-2} \\ =\underset{x\to 2}{\lim }\frac{(4x-3)(x-2)}{x-2} \\ =\underset{x\to 2}{\lim }(4x-3) \\ =4\left(2\right)-3 \\ =5 \end{gathered}$$

Part(b) Step 1. Explanation

We know,

$$\begin{gathered} m_{\text{tan}}=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c} \\ wherem=slope. \end{gathered}$$

On substituting the point$(2,-9)$

$$\begin{gathered} m_{\text{tan}}=\underset{x\to 2}{\lim }\frac{4x^2-11x-3-(-9)}{x-2} \\ =\underset{x\to 2}{\lim }\frac{4x^2-11x+6}{x-2} \\ =\underset{x\to 2}{\lim }\frac{(4x-3)(x-2)}{x-2} \\ =\underset{x\to 2}{\lim }4x-\underset{x\to 2}{\lim }3 \\ =5 \end{gathered}$$

The equation of a tangent line is,

$y-f\left(c\right)=m_{\tan }(x-c)$

On substituting,

$$\begin{gathered} y-(-9)=5(x-2) \\ y+9=5x-10 \\ y=5x-19 \end{gathered}$$

Part(c) Step 1. Explanation

The equation of the tangent line for the function

$f\left(x\right)=4x^2-11x-3$at $(2,-9)$is$y=5x-19$.

Plotting both the lines, we get the graph,