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Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,17). Express your answer in the form "ax^2+bx+c".

 Jan 4, 2022
 #1
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Since the parabola has a vertical axis of symmetry, we know that the parabola opens up or down.

We can use vertex form y=a(x-h)^2+k

 

If the vertex is (h,k) then we can substitute to find:

y=a(x-2)^2+4

We need to find a and we have only 1 equation, so we can use the other point for x and y:

17=a(1-2)^2+4

17=a+4

a=13

 

The graph of the parabola in vertex form would be y=13(x-2)^2+4

Converting to Standard Form, we get:

y=13(x-2)^2+4

y=13(x^2-4x+4)+4

y=13x^2-52x+52+4

y=13x^2-52x+56

 Jan 4, 2022
 #2
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Vertex (2,4)  form of parabola

 

y = a (x -2)^2 + 4     sub in the point  (1,17) to calculate 'a'

17 = a (1-2)^2 + 4    so a = 13

 

y = 13 ( x-2)^2 + 4   expand

y = 13x^2 -52x + 56     

 Jan 4, 2022

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