SOLUTION: Find the rule of a quadratic function if it has a minimum value of y=4, an axis of symmetry at x=3 and passes through point (4,-3)

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Question 939999: Find the rule of a quadratic function if it has a minimum value of y=4, an axis of symmetry at x=3 and passes through point (4,-3)
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
EDITING:

How I knew that? Standard Form for a parabola equation.
y=a%28x-h%29%5E2%2Bk, standard form.
Vertex is the point (h,k).

BETTER------------------------------------

Standard Form, y=a(x-h)^2+k for extreme point (h,k).
You have symmetry axis on x=3 and mainimum value y at y=4.
You know now your (h,k) minimum vertex is (3,4).

Find the factor, a.
y-k=a%28x-h%29%5E2
%28y-k%29%2F%28x-h%29%5E2=a
Substituting for h, k, and the known given point (4,-3)
Get the value for a:
a=%28-3-4%29%2F%284-3%29%5E2
a=-7%2F1%5E2
a=-7

Equation is highlight%28y=-7%28x-3%29%5E2%2B4%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Find the rule of a quadratic function if it has a minimum value of y=4, an axis of symmetry at x=3 and passes through point (4,-3)

Vertex form of a parabolic equation: y+=+a%28x+-+h%29%5E2+%2B+k
With:
x being 4
y being - 3
h being 3, and
k being 4, this becomes: -+3+=+a%284+-+3%29%5E2+%2B+4
- 3 = a + 4
a = - 3 - 4, or - 7
Rule, or equation: highlight_green%28y+=+-+7%28x+-+3%29%5E2+%2B+4%29
This is IMPOSSIBLE. For a parabola to have a vertex of (3, 4) and pass through the point, (4, - 3), it
WILL HAVE a MAXIMUM, not a MINIMUM. If it does have a MINIMUM at (3, 4), it will open UPWARDS, and therefore,
will NEVER pass through the point (4, - 3), which by the way is a point in the 2nd quadrant. In other words,
its range would be y+%3E=+4
The above equation represents what the problem states, with the exception that the graph will have a MAXIMA
instead of a MINIMA, as stated before.