One of the best uses of differentiation is to find the gradient of a point along the curve. This can help us sketch complicated functions by find turning points, points of inflection or local min or maxes.

If we look at the function $y=\frac{1}{2}x-\frac{3}{2}+\frac{17}{4}$

It’s hard to see immediately how this curve will look just by looking at the function. However, we can recognise that this function is of the form y = ax2 + bx + c meaning that it’s a quadratic equation and can be graphed as some form of parabola.

To find y intercept let x = 0 and solve the equation

In order to find the y intercept we set x equal to zero.

$y=\frac{1}{4} \times 0 - \frac{3}{2} \times 0 +\frac{17}{4}\newline y=\frac{17}{4}$

To find x intercept let y = 0

It is clear that any attempt to find an x-intercept will be complicated.

To find out if our equation does actually touch the x axis we can use the discriminant, which you hopefully have at least seen before.

$b^{2}-4ac>0$

In this formula b, a & c refers to equation(y = ax2 + bx + c. A quadratic will have x intercepts if $b^{2}-4ac>0$. For our case:

$\frac{3}{2}^{2}-4\times \frac{1}{4} \times \frac{17}{4}=-\frac{8}{4}= -2$

As such, our function will not have any x intercepts as when we substitute the terms from our function into  we get -2.

There is an easy way through differentiation to find a turning point for this function. First we take a derivative, using power differentiation.

$\frac{dy}{dx}=2\times \frac{1}{4} x^{2-1}-1 \times \frac{3}{2} \times x^{1-1} \newline \frac{dy}{dx}=\frac{1}{2} x-\frac{3}{2}$

This gives us the gradient function of the original function, so if we sub in any value of x at any of these points then we get the gradient at that point.

To find the x intercept of the gradient function we set dy/dx to 0.

To find the x intercept of the gradient function we make dy/dx = 0

$\frac{dy}{dx}=\frac{1}{2} x-\frac{3}{2}=0 \newline \frac{1}{2} x=\frac{3}{2} \newline x=3$

We can verify this by drawing a graph of the gradient function, which we can see that y intercept is negative 1.5 and the slope is 0.5

We can see that this graph describes the gradient of the original function as starting off decreasing at a large amount but at a slowing rate. When x=3 we can see that gradient equals 0 and then the curve’s gradient starts increasing towards infinity. This is consistent with what we would expect a quadratic to be.

So what does this mean? The key point to take away is that when a gradient equals 0 it is the sign of either a turning point or a point of inflection. So for our function above we can see that there is a turning point (as it is a quadratic it cannot be a point of inflection) at x=3.

If we substitute x = 3 back into the original equation:

\$latex y(3)=\frac{1}{4} \times 3 – \frac{3}{2} \times 3 +\frac{17}{4}\newline y=2

So our y co-ordinate of the turning point is 2

Using the y intercept of 17/4 and the turning point of x=3 and y=2, means we can sketch the graph and compare it to our calculated gradient.

In the graph we can see that when x is less than 3, the gradient line continually decreases and the gradient of the quadratic does the same. When x is greater than three the reverse occurs and as the gradient line increases, the gradient of the quadratic also increases.

This shows the importance of calculating the derivative and setting it to 0 to calculate turning points and points of inflection. If you are not feeling confident on any of these sketching techniques listed above have a look at our videos and pages in the functions and graphs study area.