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a curve has equation y = f(x). (a) write an expression for the slope of…

Question

a curve has equation y = f(x). (a) write an expression for the slope of the secant line through the points p(8, f(8)) and q(x, f(x)). options: $\frac{f(8)-8}{f(x)-x}$, $\frac{x - 8}{f(x)-f(8)}$, $\frac{f(x)-x}{f(8)-8}$, $\frac{f(x)-f(8)}{x - 8}$ (correct). (b) write an expression for the slope of the tangent line at p. options: $lim_{x
ightarrow0}\frac{f(x)-x}{f(8)-8}$, $lim_{x
ightarrow8}\frac{x - 8}{f(x)-f(8)}$, $lim_{x
ightarrow0}\frac{f(x)-f(8)}{x - 8}$, $lim_{x
ightarrow8}\frac{f(x)-f(8)}{x - 8}$ (incorrect)

Explanation:

Step1: Recall slope - formula for secant line

The slope of a line through two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For points $P(8,f(8))$ and $Q(x,f(x))$, we have $x_1 = 8,y_1=f(8),x_2=x,y_2 = f(x)$. So the slope of the secant line is $\frac{f(x)-f(8)}{x - 8}$.

Step2: Recall slope - formula for tangent line

The slope of the tangent line at a point $P(x_0,y_0)$ on the curve $y = f(x)$ is the limit of the slopes of the secant lines as the second - point approaches the first - point. Here $x_0 = 8,y_0=f(8)$. The slope of the secant line through $P(8,f(8))$ and $Q(x,f(x))$ is $\frac{f(x)-f(8)}{x - 8}$, and the slope of the tangent line at $P$ is $\lim_{x
ightarrow8}\frac{f(x)-f(8)}{x - 8}$.

Answer:

(a) $\frac{f(x)-f(8)}{x - 8}$
(b) $\lim_{x
ightarrow8}\frac{f(x)-f(8)}{x - 8}$