Given that the graph of f(x) passes through the point (6,3) and that the slope of its tangent line at (x,f(x)) is 2x+6, what is f(5)?

d f/dx=2x+6

f(x)=x^2+6x + k
3=6^2+6*6+k
k= -72+3=-69
f(5)=5^2+6*5-69
f(5)=25+30-69= ... you do it.

f'(x)=2x+6

Integrate to get the graph's function:
f(x)=x^2+6x+c
where c is the constant of integration.
Sub in the pt (6,3) to solve for c and hence have a fully-defined function for the graph.
Finally, sub x=5 into the function to find f(5).

To find the value of f(5), we can use the information given about the slope of the tangent line at any point on the graph.

The slope of the tangent line at (x, f(x)) is given as 2x + 6. This means that the derivative of the function f(x) is 2x + 6.

To find the function f(x), we need to integrate the derivative 2x + 6.

∫(2x + 6) dx = x^2 + 6x + C

Where C is the constant of integration.

Now, we can use the point (6, 3) to find the value of C.

Substitute x = 6 and f(x) = 3 into the equation:

6^2 + 6(6) + C = 3
36 + 36 + C = 3
72 + C = 3
C = -69

So, the function f(x) is given by:

f(x) = x^2 + 6x - 69

To find f(5), substitute x = 5 into the function:

f(5) = 5^2 + 6(5) - 69
f(5) = 25 + 30 - 69
f(5) = 55 - 69
f(5) = -14

Therefore, f(5) = -14.