a. use the definition mtan = lim x→a (f(x) - f(a))/(x - a) to find the slope of the line tangent to the…

a. use the definition mtan = lim x→a (f(x) - f(a))/(x - a) to find the slope of the line tangent to the graph of f at p. b. determine an equation of the tangent line at p. c. plot the graph of f and the tangent line at p. f(x)=x² - 4, p(3,5) a. mtan = □

a. use the definition mtan = lim x→a (f(x) - f(a))/(x - a) to find the slope of the line tangent to the graph of f at p. b. determine an equation of the tangent line at p. c. plot the graph of f and the tangent line at p. f(x)=x² - 4, p(3,5) a. mtan = □

Answer

Answer:

a. $m_{tan}=6$ b. $y = 6x - 13$

Explanation:

Step1: Identify the function and point values

$f(x)=x^{2}-4$, $a = 3$, $f(a)=f(3)=3^{2}-4=5$

Step2: Calculate the slope using the limit - definition

[ \begin{align*} m_{tan}&=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}\ &=\lim_{x\rightarrow3}\frac{(x^{2}-4)-(3^{2}-4)}{x - 3}\ &=\lim_{x\rightarrow3}\frac{x^{2}-4 - 5}{x - 3}\ &=\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}\ &=\lim_{x\rightarrow3}\frac{(x + 3)(x - 3)}{x - 3}\ &=\lim_{x\rightarrow3}(x + 3)\ &=3+3\ &=6 \end{align*} ]

Step3: Find the equation of the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,5)$ and $m = 6$. [ \begin{align*} y-5&=6(x - 3)\ y-5&=6x-18\ y&=6x-13 \end{align*} ] (Note: For part c, plotting the graph of $y=x^{2}-4$ and $y = 6x - 13$ can be done using graphing software like Desmos or a graphing calculator. The graph of $y=x^{2}-4$ is a parabola opening upwards with vertex at $(0,-4)$ and the graph of $y = 6x - 13$ is a straight line with slope 6 and y - intercept - 13. They intersect at the point $(3,5)$.)