Find an equation of the tangent line to the graph of the function f defined by the equation at the indicated point.

Question

Quizplus · Accepted Answer

We can use the point-slope form of the equation of a line to find the equation of the tangent line at the point $(4,2)$:

$$y - f(x_0) = f'(x_0)(x - x_0)$$

where $f(x) = \sqrt{x+1}$, $x_0 = 4$, and $f'(x)$ is the derivative of $f(x)$ evaluated at $x=x_0$.

We have:

$$f'(x) = \frac{1}{2\sqrt{x+1}}$$

so

$$f'(4) = \frac{1}{2\sqrt{5}}$$

Substituting everything into the point-slope equation of the tangent line, we get:

$$y - 2 = \frac{1}{2\sqrt{5}}(x - 4)$$

Simplifying, we get:

$$y = \frac{1}{2\sqrt{5}}x + \frac{3}{\sqrt{5}}$$

Therefore, the answer is $\boxed{	extbf{(B)}}$.

Find an Equation of the Tangent Line to the Graph