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.

Question

Please help.

Answer 1

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?

Answer 2

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.