Answer
The tangent line to the curve at $x=2$ is $y=\frac{1}{2}x+2$.
Work Step by Step
$$y=f(x)=\sqrt{x^2-x+7}=(x^2-x+7)^{1/2}$$
a) Find $f'(x)$:
Using the Chain Rule: $$f'(x)=\frac{1}{2}(x^2-x+7)^{-1/2}(x^2-x+7)'=\frac{2x-1}{2\sqrt{x^2-x+7}}$$
b) For $x=2$, we have $$f(2)=\sqrt{2^2-2+7}=\sqrt{9}=3$$
$$f'(2)=\frac{2\times2-1}{2\sqrt{2^2-2+7}}=\frac{3}{2\sqrt9}=\frac{3}{2\times3}=\frac{1}{2}$$
As $f'(2)$ is the slope of the tangent line to $f(x)$ at $x=2$, we can find the tangent to $f(x)$ at $x=2$ following the formula:
$$y-f(2)=f'(2)(x-2)$$ $$y-3=\frac{1}{2}(x-2)$$ $$y-3=\frac{1}{2}x-1$$ $$y=\frac{1}{2}x+2$$
