Answer
$$f(1)=8$$
Work Step by Step
By the definition of the equation of the tangent line, the slope of that line is $f'(x)$ so:
$$f'(x)=3-4x$$
Integrate both sides with respect to $x$:
$$\int f'(x)dx=\int (3-4x)dx$$
$$f(x)=3x-4\cdot \frac{x^{2}}{2}+c$$
$$f(x)=3x-2x^{2}+c$$
Use the condition $f(2)=5$ to find $c$.
$$f(2)=3\cdot 2-2\cdot 2^{2}+c$$
$$5=3\cdot 2-2\cdot 2^{2}+c$$
$$5=6-8+c$$
$$5=-2+c$$
$$7=c$$
So
$$f(x)=3x-2x^{2}+7$$
$$f(1)=3\cdot 1-2\cdot 1^{2}+7=3-2+7=8$$
