#### Quadratic gradients

For each of the following questions, describe what you think will happen before you try it out on the interactivity. Then reflect on your ideas: did the actual behaviour surprise you? And can you explain the observed behaviour? (You might want to start with very simple values for \(a\), \(b\) and \(c\).)

What happens to the gradients as you change \(c\)?

What happens to the gradients as you change \(a\)?

- What happens to the gradients as you change \(b\)?

As we change \(c\), the curve just translates up or down, so the gradient at a given \(x\)-value does not change.

The graph of the gradients is a straight line, and its gradient changes as we change \(a\).

The gradient function of \(y=x^2\) is \(2x\), as we discovered in Zooming in. If we stretch this function in the \(y\)-direction by a factor of \(2\), by setting \(a=2\), we double the gradients, giving a gradient function of \(4x\).

Does the same sort of thing happen for other values of \(a\)?

Does changing \(b\) affect the gradient of the gradient function line?

Changing \(b\) translates the gradient function. If we add \(1\) to \(b\), then the gradient function increases by \(1\). This makes some sense: \(b=1\) (with \(a=c=0\)) gives the function \(y=x\), which has gradient \(1\) everywhere. So if we add \(x\) to our function, it seems reasonable that the gradient should increase by \(1\). For example, when we go from \(y=3x\) to \(y=3x+x=4x\), the gradient increases from \(3\) to \(4\).

Putting these together, to find the gradient of the quadratic function \(f(x)=ax^2+bx+c\), we can start with \(x^2\), stretch it by a factor of \(a\) to get \(ax^2\) (which has what gradient function?), then add on \(bx\) (which does what to the gradient function?) and then finally add \(c\). This will give us the final gradient function.

#### Cubic gradients

Following on from the above ideas, we can predict what will happen for a cubic: starting with \(x^3\), which appears to have a quadratic gradient function (which quadratic exactly?), we stretch it to obtain \(ax^3\), and this stretches the gradient function. We then add on extra terms, whose behaviour we now understand from exploring quadratics: first we add on \(bx^2\), then \(cx\) and finally \(d\).

Using this idea, we should be able to predict the gradient function of any cubic \(f(x)=ax^3+bx^2+cx+d\).

Can you predict what the gradient will be for quartics (equations of the form \(y=ax^4+bx^3+\cdots\)) or polynomials of higher degree?

Some patterns definitely seem to be emerging here! What would you expect the gradient function of \(x^4\) to be?