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###### Exercises

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Exercises 1 | |

Exercises 2 | |

Exercises 3 Let's start by recalling the vertex form of a quadratic function. f(x)=a(x−h)2+k In this form the vertex of the parabola is the point (h,k), and the axis of symmetry is the vertical line x=h. Now consider the given function. p(x)=21(x+1)2−1⇕p(x)=21(x−(-1))2+(-1) We can see that h=-1 and that k=-1. Therefore, the vertex is (-1,-1), and the axis of symmetry is x=-1. To graph the function we will make a table of values. Make sure to include x-values to the left and to the right of the axis of symmetry.x21(x+1)2−1p(x)=21(x+1)2−1 -321(-3+1)2−11 -221(-2+1)2−1-21 021(0+1)2−1-21 121(1+1)2−11 Let's now draw the parabola that connects the obtained points and the vertex. We will also draw the axis of symmetry x=-1, and the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. The graphs do not have the same axis of symmetry. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=-1. The graph of the given function is wider than the graph of the parent function. The vertex of the given function, (-1,-1), is below and to the left of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of p is a vertical shrink by a factor of 21, a vertical translation down 1 unit, and a horizontal translation left 1 unit of the graph of f. | |

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Exercises 9 | |

Exercises 10 | |

Exercises 11 | |

Exercises 12 | |

Exercises 13 | |

Exercises 14 | |

Exercises 15 | |

Exercises 16 | |

Exercises 17 |