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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".

 Jun 15, 2021
 #1
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The parabola is 3x^2 - 2x + 8.

 Jun 15, 2021
 #2
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Equation of vertical parabola is 

              (x - h)2 = 4a(y - k)

where (h,k) is the vertex. 

⇒ (x - 2)2 = 4a(y - 4)

 

∵ It passes through the point (1,17)

(1 - 2)2 = 4a(17- 4)

52a = 1

a = 1/52 

 

∴ Equation of parabola is 

(x - 2)2 = 4/52 (y - 4)

13(x2 - 4x + 4) = y - 4

y - 4 = 13x2 - 52x + 52

y = 13x2 - 52x + 56

It's vertex is at (2,4) and it passes through (1,17). 

You can check here if you want https://www.desmos.com/calculator/d8z6qk1mfa

 Jun 15, 2021
edited by amygdaleon305  Jun 15, 2021

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