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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`

Answer: (-3, 18)