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".
+526 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).
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