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)

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