SOLUTION: Find the quadratic function that has given vertex and goes through the given point. (5,6), point (-1,0)

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Question 1125823: Find the quadratic function that has given vertex and goes through the given point. (5,6), point (-1,0)
Found 3 solutions by MathLover1, josgarithmetic, MathTherapy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Find the quadratic function that has given vertex and goes through the given point.
vertex: (5,6),
point: (-1,0)

first, we recall a function f%28x%29=a%28x-h%29%5E2%2Bk, which will give us vertex (h,k)=(5,6)
So the function we are looking for should look like
f%28x%29=a%28x-5%29%5E2%2B6,
and since the parabola passes through (-1,0) , we have
0=a%28-1-5%29%5E2%2B6, which means
0=36a%2B6+
36a=-6
a=-6%2F36
+a=-1%2F6
and equation is:
f%28x%29+=+-%281%2F6%29%28x-5%29%5E2%2B6




Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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Find the quadratic function that has given vertex and goes through the given point. (5,6), point (-1,0)
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vertex (5,6)
through point (-1,0)

(Note that leading coefficient will be negative, and the other root is (11,0) , if choosing vertical symmetry axis).

y=a%28x%2B1%29%28x-11%29, using the two zeros;
-
6=a%285%2B1%29%285-11%29, plugging in vertex values;
6=6%2A%28-6%29a
a=-1%2F6

highlight%28y=-%281%2F6%29%28x%2B1%29%28x-11%29%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Find the quadratic function that has given vertex and goes through the given point. (5,6), point (-1,0)
Vertex form of a parabolic function: matrix%281%2C3%2C+f%28x%29%2C+%22=%22%2C+a%28x+-+h%29%5E2+%2B+k%29

matrix%281%2C3%2C+0%2C+%22=%22%2C+a%28-+1+-+5%29%5E2+%2B+6%29 -------- Substituting (- 1, 0) for [x, f(x)] and (5, 6) for (h, k)

This should give you an "a" value of -+1%2F6, and the equation: highlight_green%28matrix%281%2C3%2C+y%2C+%22=%22%2C+%28-+1%2F6%29%28x+-+5%29%5E2+%2B+6%29%29

That's IT!!