Let's look at the graph of a typical polynomial function, of a kind you have seen before.
We'll choose a simple cubic function:

  y = x3 - 2x2 - 5x + 6

 You have learned how to graph this function using the y-intercept  (0 , 6)  and by finding its zeros, using the Factor Theorem; when you factor the polynomial, you get:
y = (x - 1)(x - 3)(x + 2)
We now have a y-intercept of  (0 , 6)  and zeros  (1 , 0),  (3 , 0)   and  (-2 , 0)

This is the graph that results:

Now lets look at a simple method you can use to identify exactly the local maximum and minimum points of this graph. Specifically, what are the points at  A and  B?

In order to do this, we are going to differentiate the polynomial function   y = x3 - 2x2 - 5x + 6.
We want to find the first derivative of this function. Since it's a polynomial, we can do this by finding the first derivative of each term.

The procedure is simple: for each term, bring the exponent out front and multiply by any coefficient already there. The remaining exponent will be one less than it was before.

Let's differentiate the polynomial, term by term:

  x3  becomes  3x2

-2x2  becomes  -4x1

-5x1  becomes  -5x0 or just -5

 6  (or  6x0)  becomes  0x-1 (or just  0)

So the function   y = x3 - 2x2 - 5x + 6  has the derivative function   y' = 3x2 - 4x - 5


The notation we're using for a first derivative function is  y'. There are three common notations:
We'll use the last one.

The derivative function can be thought of as the slope function for the original graph.
By using the function  y' = 3x2 - 4x - 5,  you can determine the slope at any point  x   just by filling in x.

For example, look at the graph at about the first x-intercept, or  (-2 , 0). The graph there is very steep. What do you think its slope is at that point?
We could draw the tangent at  (-2 , 0)  and find the slope of that line. But that's too much work, and not very accurate, as you've learned. Instead, let's use the first derivative function  y' = 3x2 - 4x - 5

Fill in the value  x = -2.  When you do, you will get y = 15
So the slope at  (-2 , 0)  is  15

Now back to our original problem, how to find the local maximum and minimum points of the graph.

The object is to find the points marked A and B. What is special about those points?
At points A and B, the slope of the function is zero! We've shown that here by drawing the tangents.
If we could find where the slope was zero, we'd know where A and B were.
But we have an equation for the slope of this function:
it's      y' = 3x2 - 4x - 5
All that is necessary is to let y' be zero, and solve the resulting equation.
In other words, solve:
3x2 - 4x - 5 = 0

Using the General Quadratic Formula, (since the equation won't factor), we get the following two rounded solutions :    x = -0.8   and   x = 2.1

Next, filling each of these values into the original function  y = x3 - 2x2 - 5x + 6  yields the corresponding rounded  y  values of  8.2  and  -4.1

So we now have the values where the slope is zero:  A (-0.8 , 8.2)  and  B (2.1 , -4.1)

Here's a summary of the method we used to find the local maximum and minimum points:

Differentiate the function. Set that first derivative equation to be equal to zero (because the slope is zero at the max and min points). Solve for the two 'x' values. Substitute them into the original equation to find the corresponding 'y' values.

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