Question

If the tangent line to y = f(x) at (7, 5) passes through the point (0, 4), find f(7) and f '(7).

Answer 1

As given,
"tangent line to f(x) is tangent at (7,5)",  therefore f(7)=5.

Line is tangent at (7,5) and passes through (0,4), therefore the slope of the line is the slope of f(x=7) = f'(7).
$f'(7) = slope\, of\, line = \frac{y_2-y_1}{x_2-x_1} = \frac{5-4}{7-0}=\frac{1}{7}$

Answer 2

The equation of tangent line to y=f(x) at point $(x_0,y_0)$ is

$y=f'(x_0)(x-x_0)+y_0.$

In your case the tangent point is (7,5), then $x_0=7,\ y_0=f(x_0)=f(7)=5$ and the equation of tangent line is

$y=f'(7)(x-7)+5.$

This tangent line passes through the point (0,4), then the coordinates of this point satisfy the equation:

$4=f'(7)(0-7)+5,\\ \\f'(7)\cdot (-7)=4-5,\\ \\-7f'(7)=-1,\\ \\f'(7)=\dfrac{1}{7}.$

Answer: $f(7)=5,\ f'(7)=\dfrac{1}{7}.$