use the function below to answer parts (a)-(c)\n$f(x)=x^{2}-3$\n(a) use the formal definition to find the…

use the function below to answer parts (a)-(c)\n$f(x)=x^{2}-3$\n(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 1$\n(b) find $f(1)$ and find the equation of the tangent line at the point $(1,f(1))$\n(c) graph $y = f(x)$ and the tangent line at the point $(1,f(1))$ in the same coordinate system\n(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\nof the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$\n$f(1)=\\lim_{h\\to0}\\frac{(1 + h)^{2}-3-(1)^{2}-3}{h}$\nevaluate the limit expression to find $f(1)$.\n$f(1)=2$ (type an integer or a fraction.)\n(b) $f(1)=-2$ (type an integer or a fraction.)\nthe equation of the tangent line at $(1,f(1))$ is $\\square$. (type your answer in slope - intercept form.)

use the function below to answer parts (a)-(c)\n$f(x)=x^{2}-3$\n(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 1$\n(b) find $f(1)$ and find the equation of the tangent line at the point $(1,f(1))$\n(c) graph $y = f(x)$ and the tangent line at the point $(1,f(1))$ in the same coordinate system\n(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\nof the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$\n$f(1)=\\lim_{h\\to0}\\frac{(1 + h)^{2}-3-(1)^{2}-3}{h}$\nevaluate the limit expression to find $f(1)$.\n$f(1)=2$ (type an integer or a fraction.)\n(b) $f(1)=-2$ (type an integer or a fraction.)\nthe equation of the tangent line at $(1,f(1))$ is $\\square$. (type your answer in slope - intercept form.)

Answer

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})) [ \begin{align*} y-(-2)&=2(x - 1)\ y + 2&=2x-2\ y&=2x-4 \end{align*} ]

Answer:

(y = 2x-4)