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.
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.
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)