Let f be a function such that f "(x) = 6x + 8.
Find f(x) if the graph of f is tangent to the line
3x - y = 2 at the point (0,-2).

Question

Let f be a function such that f "(x) = 6x + 8.
Find f(x) if the graph of f is tangent to the line 
3x - y = 2 at the point (0,-2).

anonymous · Answer

good lord

anonymous · Answer

If you integrate twice, you'll get 2 constants of integration. So basically you need to find those 2 constants.
f'(x)=3x^2+8x+c and
f(x)=3/2 x^3 + 4x^2 + cx +k.
To find c and k you need 2 equations.First the curve passes through (0,-2) and then f'(x) at (0,-2) is same as slope of the line.

anonymous · Answer

3x - y = 2 you need to do derivative

anonymous · Answer

$$f'(x)=3x^2+8x+c$$ and we know at x = 0 the slope is 3 (because that is the slope of the line) so set 
$$f'(0)=c=3$$ and we know 
$$c=3$$ so 
$$f'(x)=3x^2+8x+3$$

anonymous · Answer

to find y; y;;

anonymous · Answer

now we know that 
$$f(x)=x^3+4x^2+3x+c$$ and now we know that if x = 0 then y = -2, so solve 
$$f(0)=c=-2$$ and therefore 
$$c=-2$$

anonymous · Answer

thus the function must be 
$$f(x)=x^2+4x^2+3x-2$$

anonymous · Answer

AWESOME...THANK YOU SO MUCH, THIS IS WHAT I GOT AS WELL