Math Help!

A.) Determine "a" and "k" so both points are on the graph of the function.

1.) (0,1) (2,1); y=a(x-1)^2+k
2.) (1,11) (2,-19); y=a(x+1)^2+k


B.) Determine whether or not the function f(x)=0.25(2x-15)^2+15 has a mazimum or a minimum value. Then find the value.


C.) When y=2(x-3)(x+5) is written in standard form, what is the value of b?


D.) For y=3x^2-7x+5, what is the x-value of the vertex? Enter your answer as an improper fraction in simplest form.


THANK YOU SO MUCH!!!!! :)

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asked by Lanie
  1. A1
    1 = a(-1)^2 + k
    1 = a(1)^2 + k
    ----------------
    -1 = 0 + 0 no solution possible

    A2
    11 = a(2^2) + k
    -19 = a(3^2) + k
    -----------------subtract
    30 = -5 a
    a = -6
    11 = -6(4) + k
    k = 35
    so
    y = -6(x+1)^2 + 35

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    posted by Damon
  2. B.) Determine whether or not the function f(x)=0.25(2x-15)^2+15 has a mazimum or a minimum value. Then find the value.

    when x gets big + or big -, y gets big +
    SO
    This parabola has a minimum (holds water)
    it is symmetric about the point where
    2x-15 = 0
    or x = 7.5
    then calculate y at x = 7.5 to get the vertex

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    posted by Damon
  3. C.) When y=2(x-3)(x+5) is written in standard form, what is the value of b?

    y = 2(x^2 + 2 x - 15)

    y = 2 x^2 + 4 x - 30

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    posted by Damon
  4. D.) For y=3x^2-7x+5, what is the x-value of the vertex? Enter your answer as an improper fraction in simplest form.

    3 x^2 - 7 x = y - 5

    x^2 -(7/3) x = y/3 - 5/3

    x^2-(7/3)x+(7/6)^2 = y/3 -5/3+(7/6)^2

    (x-7/6)^2 = y/3 etc etc
    so
    when x = 7/6 we are at the vertex

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    posted by Damon

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