What is f(x+h) of 4x^2+3x-7

Question

F(x)=4x²+3x-7

solve f(x+h)

Answer 1

No solution to an expression,

f(x + h) = 4(x + h)² + 3(x + h) + 7

Answer 2

everywhere there's an x, replace it with x+h

to get

f(x+h) = 4(x+h)² + 3(x+h) -7. That could be expanded to

f(x+h) =4x² + 8xh +4h² +3x +3h -7

this looks like the beginning of showing how to take a derivative

where the derivative = the limit as h approaches zero of the ratio [f(x+h)-f(x)]/h

f(x+h)-f(x)=8xh+4h²+3h Divide that by h to get 8x+4h+3

then let h=0, to get 8x+3

8x+3 is the first derivative of f(x)=4x²+3x-7

If this is a precalculus question, that's what they're getting you ready to learn, how to take a derivative of a function. f'(x)=8x+3 if f(x)=4x²+3x-7

Answer 3

f(x)=4x²+3x-7

Replace x with x+h

then f(x+h)=4(x+h)² + 3(x+h) - 7

f(x+h)=4x² + 8xh + 4h² + 3x + 3h - 7