Asked by Kevin
                The parabola with the general equation y= ax^2 + bx+ 9 where0 <  a < 10 and 0 < b < 20 touches the x-axis at one point only. The graph passes through the point (1, 25). Find the values of a and b.
Please help.
            
        Please help.
Answers
                    Answered by
            MsPi_3.14159265
            
    Have you thought about narrowing it down a bit?
Suggestions...
1) sketch it with the information given (one x intercept)
2) sub in your point on the quadratic and see what results :)
What do those two suggestions help you to see?
    
Suggestions...
1) sketch it with the information given (one x intercept)
2) sub in your point on the quadratic and see what results :)
What do those two suggestions help you to see?
                    Answered by
            Reiny
            
    For the equation ax^2 + bx+ 9 = 0 to have only one solution, (the parabola touches the x-axis at one point only) , the discriminant has to be zero, that is
b^2 - 4a(9) = 0
b^2 = 36a
a = b^2/36
Also MsPi suggested subbing in the given point (1,25)
25 = a(1)^2 + b(1) + 9
a+b = 16
then b^2/36 + b = 16
b^2 + 36b - 576 = 0
b-12)(b+48) = 0
b = 12 or b = -48
but you said 0<b<20
so b = 12
ok, you finish it.
    
b^2 - 4a(9) = 0
b^2 = 36a
a = b^2/36
Also MsPi suggested subbing in the given point (1,25)
25 = a(1)^2 + b(1) + 9
a+b = 16
then b^2/36 + b = 16
b^2 + 36b - 576 = 0
b-12)(b+48) = 0
b = 12 or b = -48
but you said 0<b<20
so b = 12
ok, you finish it.
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