Analysing Graphs with Differentiation

## Analysing Graphs with Differentiation

Because differentiation gives you the gradient of a curve, you can differentiate an equation to find turning points on a graph.

A turning point has a gradient of zero, so by differentiating the equation and setting x to zero will give the x-values of any turning points. Substituting X into the original equation will give the y-value for the coordinate.

By examining the gradient on either side of a turning point, you can determine whether the turning point was a maximum, minimum or a stationary value.

## Example 1

For the equation y = x^2+4x-2, find the turning point and indicate whether it is a maximum or a minimum.

 y =x^2+4x-2 frac(text(d)y)(text(d)x) =2x + 4 Turning point is when gradient equals zero 0 =2x + 4 x =-2 Find y-coordinate y =x^2+4x-2 y =(-2)^2+4(-2)-2 y =-6 Coordinate is (-2, -6)

Check the gradient on either side of the turning point:

Set x = -3, gradient = 2(-3) + 4 = -2

Set x = -1, gradient = 2(-1) + 4 = +2

Gradient is negative (going down) then zero then positive (going up): ∖ _ ∕

This is a minimum.

Answer: (-2, -6). It is a minimum.

## Example 2

A cubic equation is given by y = x^3 + 4x^2 - 3x. What is the coordinate of the turning point where x < 0?

 y = x^3 + 4x^2 - 3x frac(text(d)y)(text(d)x) = 3x^2 + 8x - 3 Find the turning point where frac(dy)(dx)=0 0 = 3x^2 + 8x - 3 text(Use )x= frac(-b ±sqrt(b^2 - 4ac))(2a) = frac(-(8) ±sqrt((8)^2 - 4(3)(-3)))(2(3)) = -3 and frac(1)(3) Only interested in x = -3 y =(-3)^3 + 4(-3)^2 - 3(-3)  =18