Question

question 8
let $f(x) = \sqrt{18 - x}$
the slope of the tangent line to the graph of $f(x)$ at the point $(9, 3)$ is
the equation of the tangent line to the graph of $f(x)$ at $(9, 3)$ is $y = mx + b$ for
$m = $
and
$b = $
hint: the slope is given by the derivative at $x = 9$, ie.
$\left(\lim_{h \to 0} \frac{f(9+h)-f(9)}{h}\
ight)$

Explanation:

Step1: Define the difference quotient

$$m = \lim_{h \to 0} \frac{f(9+h)-f(9)}{h}$$
Substitute $f(x)=\sqrt{18-x}$:
$$m = \lim_{h \to 0} \frac{\sqrt{18-(9+h)} - \sqrt{18-9}}{h} = \lim_{h \to 0} \frac{\sqrt{9-h} - 3}{h}$$

Step2: Rationalize the numerator

Multiply numerator and denominator by $\sqrt{9-h}+3$:
$$m = \lim_{h \to 0} \frac{(\sqrt{9-h}-3)(\sqrt{9-h}+3)}{h(\sqrt{9-h}+3)} = \lim_{h \to 0} \frac{(9-h)-9}{h(\sqrt{9-h}+3)}$$
Simplify numerator:
$$m = \lim_{h \to 0} \frac{-h}{h(\sqrt{9-h}+3)}$$

Step3: Cancel and evaluate the limit

Cancel $h$ (where $h
eq 0$):
$$m = \lim_{h \to 0} \frac{-1}{\sqrt{9-h}+3}$$
Substitute $h=0$:
$$m = \frac{-1}{\sqrt{9}+3} = \frac{-1}{3+3} = -\frac{1}{6}$$

Step4: Solve for $b$ using point $(9,3)$

Use $y=mx+b$, substitute $x=9$, $y=3$, $m=-\frac{1}{6}$:
$$3 = -\frac{1}{6}(9) + b$$
Calculate $-\frac{9}{6} = -\frac{3}{2}$:
$$3 = -\frac{3}{2} + b$$
Solve for $b$:
$$b = 3 + \frac{3}{2} = \frac{9}{2}$$

Step5: Write the tangent line equation

Substitute $m=-\frac{1}{6}$ and $b=\frac{9}{2}$ into $y=mx+b$:
$$y = -\frac{1}{6}x + \frac{9}{2}$$

Answer:

The slope of the tangent line is $\boldsymbol{-\frac{1}{6}}$
$m = \boldsymbol{-\frac{1}{6}}$
$b = \boldsymbol{\frac{9}{2}}$
The equation of the tangent line is $\boldsymbol{y = -\frac{1}{6}x + \frac{9}{2}}$