LEVEL 5 MATH FUCTIONS REVIEW SOLUTIONS |

1) the curve of has vert asy: *x = 20* and horiz asy: *y = 1000*

We need to find *2d* where *d* is the distance between C, the center of *f(x)*

and the intersection point of *f(x)* with the line, slope = 1, through C.

Equation of the line is *(y - 1000) = (x - 20) **t** **y = x + 980**.*

Substituting this for *f(x)*, gives

(x - 20)² = 1600 *t** ** x = -20 **or** x = 60*

Substituting *x = 60* we get *y = 40*

Now, *d* is the distance between (20, 1000) and (60, 40) = 960.8 meters

At the closest point, the jets come within 1921.7 meters of each other

*(but it looks closer in the rear view mirror!)*.

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2) a) Since the zeros are -2 and 6, *h = 2* (midpoint), so *(h, k) = (2, 4)*.

*a = *slope of *VB = -1 *so* f(x) = - | x - 2 | + 4*

b)

c) - | x - 2 | + 4 ¥ 2 t* ** *- | x - 2 | ¥ - 2 -- divide both sides by -1 (reverse inequality)

| x - 2 | [ 2 t* ** *- 2 [ x - 2 [ 2 t* *0 [ x [ 4

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3) a) Slope of AB = *a* = -8 / 15 so *f(x) = -8/15 | x - 7.5 | + 4*.

b)

domain: [0, 15] | range: [0, 4] | increasing: [0, 7.5] | decreasing: [7.5, 15] |

f(x) > over domain | zeros: 0 and 15 | maximum: 4 |

c)

We compare similar triangles AEO and ACG.

We know that so, *4 - x = 2.4 meters*.

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4) a)

b) The inverse function is:

c) The plane's altitude is < 1000 meters for 15.16 seconds. (solve *h(t) < 1000)*.

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5)

vertical asymptote | horizontal asymptote | y-intercept | zero |

a) x = 1 | y = 3 | (0, 1) | (1/3, 0) |

b) x = 3.5 | y = 2 | (0, 5/7) | (5/4, 0) |

c) x = ½ | y = - ¾ | (0, 1) | (-2/3, 0) |

d) x = 1 | y = 1 | (0, - 1) | (- 1, 0) |

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6) Set x = 32 in each function, multiply by 3 for 3 years.

p_{1} = $300.38 |
p_{2} = $299.29 |
p_{3} = $286.50 |
p_{4} = $295.80 |

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7)

a) | b) 75x + 100 < 85x t x > 10 hours |

c) 80x = 75x + 100 t x = 20 hours | d) no, it's always $75 plus something. |

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8)

a) g [f(x) = g(x + 1) = x² - 4 | b) h [f(x)] = h(x + 1) = (2x + 1)^{½} . |

c) k [f(x)] = k(x + 1) = x / (2x + 3) | d) it's a quadratic function |

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