Q. 185 - Calculus

Question

Use the definition of the derivative to find the equations of the lines described in Exercises 59-64.

The tangent line to $f\left(x\right)=x^2$ at $x=-3$

Step-by-Step Solution

Answer

The equation of tangent line is $y=2x^2+6x+9$

Step 1. Given information

Given the function $f\left(x\right)=x^2$ and the point $x=-3$

Step 2: Use the formula for equation and calculate

Calculating, we get

$$\begin{gathered} y=f\left(c\right)+f^{\text{'}}\left(c\right)(x-c) \\ y={(-3)}^2+2x(x-(-3\left)\right) \\ y=9+2x(x+3) \\ y=9+2x^2+6x \\ y=2x^2+6x+9 \end{gathered}$$