Questions: a. Use the definition (mtan =lim h rightarrow 0 fracf(a+h)-f(a)h) to find the slope of the line tangent to the graph of (f) at (P). b. Determine an equation of the tangent line at (P). (f(x)=sqrtx+99, P(1,10)) a. (mtan =) (Simplify your answer. Type an exact answer, using radicals as needed.)

Find the slope of the line tangent to the graph of \( f \) at point \( P(1, 10) \).

Using the definition of the derivative...

We apply the limit definition of the derivative:
\[ m_{\tan} = \lim_{h \rightarrow 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \rightarrow 0} \frac{\sqrt{(1+h) + 99} - \sqrt{1 + 99}}{h} = \lim_{h \rightarrow 0} \frac{\sqrt{h + 100} - 10}{h} \]

Calculating the limit...

Evaluating the limit gives us:
\[ m_{\tan} = \frac{1}{20} \]

The slope of the tangent line is \( \boxed{\frac{1}{20}} \).

Determine an equation of the tangent line at point \( P(1, 10) \).

Using the point-slope form of a line...

The point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = \frac{1}{20} \), \( x_1 = 1 \), and \( y_1 = 10 \):
\[ y - 10 = \frac{1}{20}(x - 1) \]

Rearranging to find the equation...

This simplifies to:
\[ y = \frac{1}{20}x + \frac{199}{20} \]

The equation of the tangent line is \( \boxed{y = \frac{1}{20}x + \frac{199}{20}} \).

The slope of the tangent line is \( \boxed{\frac{1}{20}} \).
The equation of the tangent line is \( \boxed{y = \frac{1}{20}x + \frac{199}{20}} \).