Question

differentiate the function. then find an equation of the tangent line at the indicated point on the graph of the function. y = f(x)=5 + \sqrt{3 - x}, (x,y)=(-6,8)
the derivative of the function y = f(x)=5 + \sqrt{3 - x} is -\frac{1}{2\sqrt{3 - x}}.
what is the equation of the tangent line at (-6,8)?
a. y - 8 = 6(x + 6)
b. y - 8=-\frac{1}{6}(x + 6)
c. y - 8=\frac{1}{6}(x + 6)
d. y - 8=-6(x + 6)

Explanation:

Step1: Recall point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope of the line.

Step2: Find the slope of the tangent line

We are given the derivative of the function $y = f(x)=5+\sqrt{3 - x}$ as $f^\prime(x)=-\frac{1}{2\sqrt{3 - x}}$. We need to find the slope of the tangent line at the point $(x,y)=(-6,8)$. Substitute $x=-6$ into the derivative:
\[

$$\begin{align*} m&=-\frac{1}{2\sqrt{3-(-6)}}\\ &=-\frac{1}{2\sqrt{9}}\\ &=-\frac{1}{2\times3}\\ &=-\frac{1}{6} \end{align*}$$

\]

Step3: Write the equation of the tangent line

We have the point $(x_1,y_1)=(-6,8)$ and slope $m =-\frac{1}{6}$. Using the point - slope form $y - y_1=m(x - x_1)$, we get $y - 8=-\frac{1}{6}(x + 6)$.

Answer:

B. $y - 8=-\frac{1}{6}(x + 6)$