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))

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.

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:

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

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 \( m = -\frac{1}{8} \).

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 \( \frac{9}{2} \).

Final Answer