use the function below to answer parts (a)-(c).\n$f(x)=x^{2}-4$\n(a) use the formal definition to find the…

use the function below to answer parts (a)-(c).\n$f(x)=x^{2}-4$\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 $f(x)=lim _{h \rightarrow 0} \frac{\text { provided that the limit exists. use the }}{h}$\ndefinition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$\n$$f^{prime}(1)=lim _{h \rightarrow 0} \frac{left(1+h)^{2}-4\right-left(1)^{2}-4\right}{h}$$\nevaluate the limit expression to find $f^{prime}(1)$.\n$$f^{prime}(1)=2$$ (type an integer or a fraction.)\n(b) $f(1)=-3$ (type an integer or a fraction.)\nthe equation of the tangent line at $(1, f(1))$ is (type your answer in slope - intercept form.)

use the function below to answer parts (a)-(c).\n$f(x)=x^{2}-4$\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 $f(x)=lim _{h \rightarrow 0} \frac{\text { provided that the limit exists. use the }}{h}$\ndefinition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 1$\n$$f^{prime}(1)=lim _{h \rightarrow 0} \frac{left(1+h)^{2}-4\right-left(1)^{2}-4\right}{h}$$\nevaluate the limit expression to find $f^{prime}(1)$.\n$$f^{prime}(1)=2$$ (type an integer or a fraction.)\n(b) $f(1)=-3$ (type an integer or a fraction.)\nthe equation of the tangent line at $(1, f(1))$ is (type your answer in slope - intercept form.)

Answer

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)