Say we have the function with equation,

f(x) = x^{2}

We may already know what this equation looks like on a graph (a parabola) but lets proceed with steps 1-4 to determine what the curve will look like,

We do this by differentiating the function f(x) = x^{2}, then equating it to zero, and finding the values of x that make the derivative equal to zero.

So,

f(x) = x^{2}

f’(x) = 2x

Now, we want to find the flat parts of the equation, the flat parts are the points where the gradient is equal to zero. So, we equate f’(x) = 2x to zero, then we solve to find the value/s of x that make the derivative equal zero.

2x = 0

x = 0 is the only point where the gradient of the function equals zero, and is therefore our one and only flat point.

We also want to know what the y-value is for this flat point, so we input x = 0 into the original function.

y = f(0) = (0)^{2} = 0

So, our flat point is at (0, 0)

Now we have found our flat point, we can move on to step two,

To recall, the points “around” the flat points is a rough calculation and its purpose is to let us know what the curve of the gradient is before and after our flat point, this will help us get and idea of what the curve will look like,

So, as we only have one flat point, at x = 0, we can pick any value of *x* to the left and any to the right, but for this, lets choose x = -1 as our left point and x = 1 as our right point. To recall from our previous notes, if we have ** a** as our flat point, then the point to the left is

We have chosen our left and right x-values for the points “around” our flat points, x = -1 and x = 1, input these x-values into our original function to give us their y-value as well.

For, x = - 1, our y-value will be,

y = f(-1) = (-1)^{2} = 1,

So, for our point to the left, we have (-1, 1)

For, x = 1, our y-value will be,

y = f(1) = (1)^{2} = 1,

So, for our point to the right, we have (1, 1)

We have identified our points “around” our flat point, now we need to determine what the gradient of those points are. We do this by inputting the x-value of our left or right point into the derivative of our function, f’(x). Then, we look at whether the gradient is positive or negative in value.

For the point to the left, x = - 1

f’(x) = 2x

f’(-1) = 2(-1) = -2

So for our left point, we have the gradient equal to a negative number, this means the gradient is going downwards at this point.

For the point to the right, x = 1

f’(x) = 2x

f’(1) = 2(1) = 2

So for our right point, we have the gradient equal to a positive number, this means the gradient is going upwards at this point.

Now for our new Step, Step Four,

Here, we simply draw a straight upwards, downwards or flat beside/beneath our identified flat point, left point and right point. You can lay this out in a variety of ways, but below is the one we think is the best.

Using this table, the shape of the graph may start to become more visual.

That is the work we do for steps 1 to 4, now lastly the easier step of “joining the lines”

First, plot the points (-1, 1), (0, 0) and (1, 1) on plain or graph paper,

Then, look back at step 4, and roughly sketch the straight line representing the positive upwards gradient or negative downwards gradient or flat zero gradient over each of the corresponding points,

Lastly, connect the straight lines together, to get a rough idea of what the curve would look like, we may be familiar with the shape, (a parabola)

Question: Roughly sketch the curve of function f(x) = x^{3} - 3x^{2} - 24x + 7 on to a x and y graph, follow the 5 steps listed above.

The "flat" parts are the points where our function has a "flat" gradient. To find these we determine the derivative, equate it to zero, and solve for x.

f(x) = x^{3} - 3x^{2} - 24x + 7

f '(x) = 3x^{2} - 6x - 24

= 3(x^{2} - 2x - 8)

Now equate to zero and solve for *x*.

= 3(x^{2} - 2x - 8)

We can factorize this.

= 3(x + 2)(x - 4)

Now there are two parts of the function that could make the whole function equal to zero. If either of (x + 2) or (x - 4) equals zero, then the whole function equals zero. We need to equate each of the zero and solve.

(x + 2) = 0

x = -2

(x - 4) = 0

x = 4

So we have two x-values that equate the derivative to zero and therefore two points that have a gradient equal to zero ( i.e. "flat"). Now we will find their y-values.

Input x = -2 into f(x) to find its corresponding y-value.

f(-2) = (-2)^{3} - 3(-2)^{2} - 24(-2) + 7

= -8 -12 + 48 + 7

= 35

So our first flat point is (-2, 35)

Input x = 4 into f(x) to find its corresponding y-value.

f(4) = (4)^{3} - 3(4)^{2} - 24(4) + 7

= 64 - 48 - 96 + 7

= -73

So our first flat point is (-4, -73)

This simply involves picking a point to the left and to the right of each flat point with regards to the x-axis.

For flat point (-2, 35) - for the point to the left with regards to the x-axis, pick x = -3

So y = f(-3) = (-3)^{3} - 3(-3)^{2} - 24(-3) + 7

= -27 - 27 + 72 + 7

= 25

Our point to the left of (-2, 35) is (-3, 25)

We can also use the point to the right of (-2, 35) as the point to the left of our other flat point, (4, -73).

So, for our point to the right of (-2, 35) and to the left of (4, -73) with regards to the x-axis, we can pick anything greater than x = -2 and less than x = 4. Let's choose x = 1.

So, for the y-value:

y = f(1) = (1)^{3} - 3(1)^{2} - 24(1) + 7

= 1 - 3 - 24 + 7

= -19

So we have (1, -19)

Lastly, for our point to the right of (4, -73), let's choose x = 5.

So, for the y-value:

y = f(5) = (5)^{3} - 3(5)^{2} - 24(5) + 7

= 125 - 75 - 120 + 7

= -63

Gives us (5, -63)

So now we have:

Flat points; (-2, 35) and (4, -73)

And the points "around" th flat points; (-3, 25); (1, -19) and (5, -63)

This simply involves putting the x-value of the points "around" our flat points into the derivative to see if we have a positive or negative gradient.

For (-3, 25) input x = -3 into the derivative, f '(x).

f '(x) = 3(x + 2)(x - 4)

f '(-3) = 3(-3 + 2)(-3 - 4)

= 3(-1)(-7) = 21

= a positive gradient (upwards slope)

A way to do this step is to layout a table, in x-axis order, to help start visualizing what the curve of the function will look like.

The table will start to help the graph curve come together and leads to the last step which is to plot and connect these sketches.

First let's roughly plot all of our flat points and points "around" them on an x-y graph.

Now, on each point, sketch its corresponding gradient line and then connect each line.

And that is our rough graphing of a polynomial function using derivatives.