Slope: The Derivative of a Function at a Point (2nd issue)

[morrowcosom]

Homework Statement

We are calculating the slope of the function f(x) = 5 - 3x^2 at x = -1.
For the function f(x) = 5 - 3x^2, we now know:

f(-1) = 2
f(-1+h) = 5 - (3 - 6h + 3h^2)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
{f(-1+h) - f(-1)}/h
--------------------------------------------------------------------------------

Homework Equations

Slope = limh->0 [f(x0 + h) - f(x0)] / h

The Attempt at a Solution

{5-(3-6h+3h^2) -2}/h
{5-(-6h+3h^2)+1}/h
{5-3h(2-h)+1}/h
5-3(2-h)+1

I am doing independent study on a computer program and it says my solution is wrong. I have no idea how to get it in the form of 5 - 3x^2 (How to get the (2-h)^2 or how to get rid of the constant on the right, which starts out as -2 or f(-1). What am I doing wrong?

 

[Mark44]

Homework Statement

We are calculating the slope of the function f(x) = 5 - 3x^2 at x = -1.
For the function f(x) = 5 - 3x^2, we now know:

f(-1) = 2
f(-1+h) = 5 - (3 - 6h + 3h^2)

You should simplify this to 2 + 6h - 3h^2.

Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
{f(-1+h) - f(-1)}/h
--------------------------------------------------------------------------------

Homework Equations

Slope = limh->0 [f(x0 + h) - f(x0)] / h

The Attempt at a Solution

{5-(3-6h+3h^2) -2}/h
{5-(-6h+3h^2)+1}/h

The line above is incorrect. You subtracted 2 from 3 to get 1. You should have subtracted 2 from -3.

{5-3h(2-h)+1}/h
5-3(2-h)+1

I am doing independent study on a computer program and it says my solution is wrong. I have no idea how to get it in the form of 5 - 3x^2 (How to get the (2-h)^2 or how to get rid of the constant on the right, which starts out as -2 or f(-1). What am I doing wrong?