MythAsian From: US Posts: 42 Votes: 84 |
If the parabola was y=x^2+x-12 and you simplified it to (x+4)(x-3) and the x int.= -4 or 3 and the yint. = -12, what is the turning point. |

Anonymous From: - Posts: - Votes: - |
If you've done derivatives, you can find the point where the slope of the tangent is 0 dy/dx = 2x + 1 0 = 2x + 1 2x = -1 x = -1/2 and then use this value to find y from the original equation. if not, you can use the formula for the x-coordinate of the turning point x = -b/2a (from the equation y = ax^2 + bx + c) |

Anonymous From: - Posts: - Votes: - |
-1/2 Two ways to find. First it has to be half way between the x intercepts. |-4-3| =7 units. half is 3.5 add to -4 for -.5 Or take derivative of the equation y=x^2+x-12 y'=2x+1 solve for y'=0 -1=2x -1/2=xgood one idiot |

Anonymous From: - Posts: - Votes: - |
Evening |

Anonymous From: - Posts: - Votes: - |
Mcvhmkvfhkm |

Anonymous From: - Posts: - Votes: - |
Mcvhmkvfhkm |

Anonymous From: - Posts: - Votes: - |
Thanks |

Anonymous From: - Posts: - Votes: - |
Mcvhmkvfhkm |

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