Question

Please help! Will give a medal :)

Answer 1

Write a quadratic function whose graph has vertex: (3, −5) and passes through: (7, 27).

Answer 2

since the vertex is \((3,-5)\) in vertex form it looks like
\[y=a(x-3)^2+5\] the only number you do not know is \(a\)

Answer 3

ok that was a mistake, should have been \[y=a(x-3)^2-5\]

Answer 4

then since \((7,27)\) is on the graph, you know
\[27=a(7-3)^2-5\] or
\[27=16a-5\] and so you can solve for \(a\)

Answer 5

a=2

Answer 6

so it would be y=2(x-3)^2-5?

Answer 7

@satellite73

Answer 8

am I right?
@satellite73

Answer 9

y = ax^2 + bx + c = 0.
Must find 3 unknowns a, b, c by 3 equations.
a) The x- coordinate of the vertex (axis of symmetry) of the parabola:
x = -b/2a = 3 -> b = -6a (1)
b) The y-coordinate of the vertex is (-5), when x = 3:
-5 = 9a + 3b + c . Replace b by -6a
9a + 3b + c = 9a - 18a +c = = -9a + c = -5 -> c = 9a - 5.(2)
c) Write that the parabola passes at point (7, 27):
27 = 49a + (-6a)(7) + (9a - 5) = 49a - 42a + 9a - 5 = 16a - 5 = 27 (3)
16a = 32 -> a = .... -> b = .... and c = ....