How would you find the quadratic that goes through the point (4,0), the axis of symmetry is x=7, and it also goes through the point (1,5)?

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  y = (5/27)(x -7)^2 -5/3

Step-by-step explanation:

Use the given points to find the unknowns in the equation.

If the axis of symmetry is x=7, then the equation can be written in the form ...

  y = a(x -7)^2 +b

Filling in the two point values, we have two equations:

  0 = a(4 -7)^2 +b   ⇒   9a +b = 0

  5 = a(1 -7)^2 +b   ⇒   36a +b = 5

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Subtracting the first equation from the second, we have ...

  (36a +b) -(9a +b) = (5) -(0)

  27a = 5

  a = 5/27

Substituting that value into the first equation gives ...

  9(5/27) +b = 0

  5/3 +b = 0

  b = -5/3

So, the quadratic can be written in vertex form as ...

  y = (5/27)(x -7)^2 -5/3

146%

step-by-step explanation:

1.true 2.false? ? i think. .

step-by-step explanation: