SOLUTION: Find the quadratic equation with a maximum value of 10 and x-intercepts of 1 and 3
Algebra.Com
Question 378471: Find the quadratic equation with a maximum value of 10 and x-intercepts of 1 and 3
Found 2 solutions by solver91311, robertb:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I can't quite do exactly what you ask. A quadratic equation is something of the form:
which has two roots, but does NOT have a maximum value or intercepts of any kind.
On the other hand, a quadratic function is a thing of the form:
\ =\ ax^2\ +\ bx\ +\ c)
And DOES have a maximum/minimum at the vertex, x intercepts (sometimes, provided 
), a y-intercept, and an axis of symmetry. I'm going to proceed on the assumption that is what you meant in the first place.
You have specified 3 points through which the graph of the function passes. First, we have the two intercepts, )
and )
. By symmetry, the vertex, (the maximum point in this case) must lie midway between the intercepts. Hence the 
-coordinate of the vertex must be 2. And since the value of the function at the maximum is 10, the vertex must be at )
Using this data we can create three linear equations:
\ =\ a(1)^2\ +\ b(1)\ +\ c\ =\ 0)
\ =\ a(2)^2\ +\ b(2)\ +\ c\ =\ 10)
\ =\ a(3)^2\ +\ b(3)\ +\ c\ =\ 0)
which is to say:
Solve the system for 
, 
, and 
to get the coefficients of your desired quadratic function.
John
My calculator said it, I believe it, that settles it
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
The y-coordinate of the vertex is 10. To get the x-coordinate, just get the average of the x-intercepts, which is (1+3)/2 = 2. Hence the vertex is (2, 10).
From the vertex form of the equation of the parabola, we get . We have to solve for a. Then using either one of the x-intercepts (I will use (1,0)) we get
;
, or
a = -10.
Therefore the equation is .
=D
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