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Question
A graph of a parabola has line of symmetry x=3 and contains the points (5,-3) and (-1,9). determine an equation for the parabola in the form a(x-h)^2+k
Step by step answer please.
Thank you
Look at y = a(x-h)² + k.
For this parabola, it can be seen that the minimum would be at
x = h, where (x-h)² = 0. All other values of x would make
(x-h)² positive. This says the parabola is centerd at x = h.
From that, it can be said that h = 3.
The equation is now y = a(x-3)² + k.
The first point, (5,-3), gives us one equation:
-3 = a(5-3)² + k. Since 5-3 = 2, squaring it gives 4.
This makes the equation -3 = 4a + k.
The second point, (-1,9), gives us another equation:
9 = a(-1-3)² + k. Since -1 - 3 = -4, and -4² = 16,
the equation is 9 = 16a + k.
The two equations are
-3 = 4a + k
9 = 16a + k.
Solving these two equation will give you a and k.
The variable h was already given, so now you have the equation.
Hint:
To start to solve, just take eq #2 - eq#1. That will cancels the k
and leave 12 = 12a. That tells you what a is, and then k can be found. Once you have these two, you have all three, and there's the parabola.
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