SOLUTION: I need help When the range of x determined by {{{-ax^2+bx+4>=0}}} is {{{-1/3<=x<=4}}} then a = ?, b = ?

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Question 1040224: I need help
When the range of x determined by
is
then a = ?, b = ?
Found 2 solutions by rothauserc, Edwin McCravy:
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
(-a/9) - (b/3) + 4 = 0
-a -3b + 36 = 0
a ÷ 3b - 36 = 0
:
-16a + 4b + 4 = 0
:
We have two equations in 2 unknowns
:
a +3b - 36 = 0
16a - 4b - 4 = 0
:
Solve first equation for a
:
a = -3b + 36
:
Substitute for a in second equation
:
16(-3b+36) -4b -4 = 0
:
-48b +576 -4b -4 = 0
-52b = -572
b = 11
a +33 -36 = 0
a = 3
:
*******
a = 3
b = 11
*******
:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
When the range of x determined by
is
then a = ?, b = ?
In order for the function 



to be greater than or equal to 0, its graph must be on or above the
x-axis.  A quadratic equation is the equation of a parabola.  

So this is a parabola that opens downward
and has x-intercepts (real zeros) at -1/3 and 4.





So we are to find the equation of the function



In order for it to have zeros -1/3 and 4,
it must be of the form

 







We compare it to



The last (constant) terms must be the same, so





Substituting in









So a = 3 and b = 11

Edwin
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