Question

use the function below to answer parts (a)-(c)
$f(x)=x^{2}-3$
(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 $f(x)=\lim_{h\to0}\frac{}{h}$ provided that the limit exists. use the
of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$
$f(1)=\lim_{h\to0}\frac{(1 + h)^{2}-3-(1)^{2}-3}{h}$
evaluate the limit expression to find $f(1)$.
$f(1)=2$ (type an integer or a fraction.)
(b) $f(1)=-2$ (type an integer or a fraction.)
the equation of the tangent line at $(1,f(1))$ is $\square$. (type your answer in slope - intercept form.)

Explanation:

Step1: Find \(f(1)\)

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

Step2: Recall the point - slope form of a line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})\) is a point on the line and \(m\) is the slope.
We know from part (a) that the slope \(m = f^{\prime}(1)=2\) and the point \((x_{1},y_{1})=(1,f(1))=(1,-2)\)

Step3: Substitute into the point - slope form and simplify

Substitute \(x_{1}=1,y_{1}=-2,m = 2\) into \(y - y_{1}=m(x - x_{1})\)

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Answer:

\(y = 2x-4\)