Questions: Let f(x) = sqrt(20-x) The slope of the tangent line to the graph of f(x) at the point (4,4) is The equation of the tangent line to the graph of f(x) at (4,4) is y=mx+b for m= and b= Hint: the slope is given by the derivative at x=4, ie. (left(lim h -> 0 (f(4+h)-f(4))/hright))

Let f(x) = sqrt(20-x)
The slope of the tangent line to the graph of f(x) at the point (4,4) is
The equation of the tangent line to the graph of f(x) at (4,4) is y=mx+b for

m=

and

b=

Hint: the slope is given by the derivative at x=4, ie.

(left(lim h -> 0 (f(4+h)-f(4))/hright))

Transcript text: Let $f(x)=\sqrt{20-x}$ The slope of the tangent line to the graph of $f(x)$ at the point $(4,4)$ is $\square$ The equation of the tangent line to the graph of $f(x)$ at $(4,4)$ is $y=m x+b$ for \[ m= \] $\square$ and \[ b= \] $\square$ Hint: the slope is given by the derivative at $x=4$, ie. \[ \left(\lim _{h \rightarrow 0} \frac{f(4+h)-f(4)}{h}\right) \]

Solution

Solution Steps

To find the slope of the tangent line to the graph of $ f(x) = \sqrt{20-x} $ at the point $ (4,4) $, we need to compute the derivative of $ f(x) $ and evaluate it at $ x = 4 $. The derivative, $ f'(x) $, represents the slope of the tangent line at any point $ x $. Once we have the slope $ m $, we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the point of tangency.

Step 1: Find the Derivative of $ f(x) $

To find the slope of the tangent line at a specific point, we first need to determine the derivative of the function $ f(x) = \sqrt{20-x} $. The derivative is given by:

\[ f'(x) = \frac{d}{dx} \left( \sqrt{20-x} \right) = -\frac{1}{2\sqrt{20-x}} \]

Step 2: Evaluate the Derivative at $ x = 4 $

Next, we evaluate the derivative at $ x = 4 $ to find the slope of the tangent line at the point $ (4, 4) $:

\[ f'(4) = -\frac{1}{2\sqrt{20-4}} = -\frac{1}{8} \]

Thus, the slope of the tangent line at $ x = 4 $ is $ m = -\frac{1}{8} $.

Step 3: Find the Equation of the Tangent Line

Using the point-slope form of a line, $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (4, 4) $ and $ m = -\frac{1}{8} $, we can find the equation of the tangent line:

\[ y - 4 = -\frac{1}{8}(x - 4) \]

Solving for $ y $, we get:

\[ y = -\frac{1}{8}x + \frac{1}{2} + 4 \]

Simplifying, the equation becomes:

\[ y = -\frac{1}{8}x + \frac{9}{2} \]

Thus, the y-intercept $ b $ is $ \frac{9}{2} $.