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)

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