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) $f1 = - 3$ (type an integer or a fraction.)
(b) $f(1) = square$, and the equation of the tangent line at $(1, f(1))$ is $y = 2x - 5$. (type your answer in slope-intercept form.)
(c) choose the correct graph below.
a.
b.
c.
d.

Explanation:

Step1: Find the derivative using the formal definition

The formal definition of the derivative is \(f^\prime(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}\). Given \(f(x)=x^{2}-4\), then \(f(x + h)=(x + h)^{2}-4=x^{2}+2xh+h^{2}-4\).

$$
LATEXBLOCK0
$$

As \(h
ightarrow0\), \(f^\prime(x)=2x\). At \(x = 1\), \(f^\prime(1)=2\times1=2\).

Step2: Find \(f(1)\)

Substitute \(x = 1\) into \(f(x)=x^{2}-4\), so \(f(1)=1^{2}-4=-3\).

Step3: Find the equation of the tangent line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(1,-3)\) and \(m = 2\).

$$
LATEXBLOCK1
$$

Answer:

(a) The derivative of \(y = f(x)\) at \(x = 1\) is \(2\). (b) \(f(1)=-3\), and the equation of the tangent line is \(y = 2x-5\). (c) To graph \(y=f(x)=x^{2}-4\) (a parabola opening upwards with vertex \((0,-4)\)) and \(y = 2x-5\) (a straight line with slope \(2\) and \(y\) - intercept \(-5\)), plot points for the parabola (e.g., when \(x = 0,y=-4\); \(x=2,y = 0\); \(x=-2,y = 0\)) and for the line (e.g., when \(x = 0,y=-5\); when \(y = 0,x=\frac{5}{2}\)). The correct graph is the one where the line \(y = 2x - 5\) touches the parabola \(y=x^{2}-4\) at the point \((1,-3)\).