Question

use the function below to answer parts (a)-(c).
$f(x)=x^{2}-4$
(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 1$
(b) find $f(1)$ and find the equation of the tangent line at the point $(1, f(1))$
(c) graph $y = f(x)$ and the tangent line at the point $(1, f(1))$ in the same coordinate system.
(a) the derivative of a function $f$ at $x$, denoted by $f(x)$, is $f(x)=lim _{h
ightarrow 0} \frac{\text { provided that the limit exists. use the }}{h}$
definition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$

$$f^{prime}(1)=lim _{h ightarrow 0} \frac{left(1+h)^{2}-4 ight-left(1)^{2}-4 ight}{h}$$

evaluate the limit expression to find $f^{prime}(1)$.
$$f^{prime}(1)=2$$ (type an integer or a fraction.)
(b) $f(1)=-3$ (type an integer or a fraction.)
the equation of the tangent line at $(1, f(1))$ is (type your answer in slope - intercept form.)

Explanation:

Step1: Calculate \(f(1)\)

Given \(f(x)=x^{2}-4\), substitute \(x = 1\) into the function.
\(f(1)=1^{2}-4=1 - 4=-3\)

Step2: Use the point - slope form \(y - y_{1}=m(x - x_{1})\)

We know that the slope \(m = f^{\prime}(1)=2\) (from part (a)), and the point \((x_{1},y_{1})=(1,f(1))=(1,-3)\)
Substitute into the point - slope form: \(y-(-3)=2(x - 1)\)

Step3: Simplify to slope - intercept form \(y=mx + b\)

Expand \(y + 3=2x-2\)
Subtract 3 from both sides: \(y=2x-2 - 3\)
\(y=2x-5\)

Answer:

The equation of the tangent line is \(y = 2x-5\)