Calculus: Early Transcendentals 8th Edition

$f'(2)=4$ and $f(2)=3$
The equation of the tangent line $l$ to the curve $y=f(x)$ at the point where $a=2$ is $$(l):y=4x-5$$ Since $f'(2)$ is the slope of the tangent line $l$, we can deduce that $f'(2)=4$ (Remember that in a line equation $y=ax+b$, $a$ is the slope of that line) The point where $a=2$ also lies in the line $l$, therefore we can use its equation to find $f(2)$. At $a=2$, we have $$f(2)=4\times2-5=3$$