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 = 6 b. y = 6x - 13 c. choose the correct answer below.

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 = 6 b. y = 6x - 13 c. choose the correct answer below.

Answer

Answer:

a. 6 b. $y = 6x-13$ c. Without seeing the actual visual content of the options A - D clearly, we cannot determine the correct graph - choice. But to plot the graph of $y = x^{2}-4$ and $y=6x - 13$:

  • For $y=x^{2}-4$, it is a parabola with vertex at $(0, - 4)$.
  • For $y = 6x-13$, it is a straight - line with slope $m = 6$ and $y$ - intercept $b=-13$. The line should pass through the point $(3,5)$ and be tangent to the parabola $y=x^{2}-4$ at that point.

Explanation:

Step1: Find $f(a)$

Given $f(x)=x^{2}-4$ and $a = 3$, then $f(a)=f(3)=3^{2}-4=9 - 4=5$.

Step2: Calculate the slope $m_{tan}$

[ \begin{align*} m_{tan}&=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x - a}\ &=\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*} ]