Home > Differentiation > Finding the turning points of a polynomial

## Finding the turning points of a polynomial

The turning points are sometimes called the stationary points or maximums and minimums.

Lets look at what happens to the gradient at a turning point

• If we consider the maximum, you can see that before it has a positive gradient and after it has a negative gradient.  During this transition it has a gradient of zero.  So at a maximum $\frac{dy}{dx}=0$
• If we consider the minimum, you can see that before it has a negative gradient and after it has a positive gradient.  During this transition it has a gradient of zero.  So at a minimum $\frac{dy}{dx}=0$
• So we can see that at a turning point (maximum or minimum) $\frac{dy}{dx}=0$

Examples

### Find the maximum of the curve $y=15-2x-x^2$

1. We need to find the gradient function $\frac{dy}{dx}$
2. We let $\frac{dy}{dx}=0$, because this is when the maximum will occur
3. We solve the equation to find the value of x where the maximum occurs
4. We substitute this value in the original equation to get the value for y

$\frac{dy}{dx}=-2-2x$

Let $\frac{dy}{dx}=0 \Rightarrow -2-2x=0 \Rightarrow x=-1$

If $x=-1 \Rightarrow y = 15-2(-1)-(-1)^2 = 16$

$\therefore$ the maximum value is $y=16$

### Find the coordinates of the turning points of $y=x^3-6x^2+9x$

1. We need to find the gradient function $\frac{dy}{dx}$
2. We let $\frac{dy}{dx}=0$, because this is when the turning points will occur
3. We solve the equation to find the values of x where the turning points occurs
4. We substitute the values into the original equation to get the values for y

$\frac{dy}{dx}=3x^2-12x+9$

Let $\frac{dy}{dx}=0 \Rightarrow 3x^2-12x+9=0 \Rightarrow 3(x^2-4x+3)=0 \Rightarrow 3(x-1)(x-3)=0$

So $x=1$ of $x=3$

If $x=1$ then $y=4$ and if $x=3$ then $y=-18$

So the coordinates of the turning points are $(1,4)$ and $(3,-18)$

### Given that the curve $y=x^2+ax-5$ has a stationary point at $x = 4$ find the value of a

1. We need to find the gradient function $\frac{dy}{dx}$
2. We let $\frac{dy}{dx}=0$, because this is when the turning points will occur and we will let $x = 4$
3. We will solve the resulting equation to find the value of a

$\frac{dy}{dx}=2x+a$

when $x=4, \frac{dy}{dx}=0 \Rightarrow 0=8+a \Rightarrow a=-8$

### Find the equation of the tangent to the curve $y=2x^3+3x-5$ at $x=1$

1. We need to find the gradient function $\frac{dy}{dx}$, because to get the equation of a line we need the gradient and a point it passes through
2. Substitute $x=1$ into the equation for y to get the required point the line passes through
3. Substitute $x=1$ into the equation for $\frac{dy}{dx}$ to get the gradient of the line
4. Use $y-y_1=m(x-x_1)$  to get the equation of the line

$\frac{dy}{dx}=6x^2+3$

At $x=1, y=0$

At $x=1, \frac{dy}{dx}=9$

So substituting into $y-y_1=m(x-x_1)$ gives $y=9(x-1)$