quadradic problem, need help?

Question

The quadratic function f(x) is negative for x>9/2 and x< -1, but no other value of x. If f(1)=28 determine f(x) algebraically and sketch a graph of the function.

Answer 1

The general form for a quadratic equation is going to be y = ax^2 + bx + c. The trick here is to find a, b, and c.

We have zeros - at x > 9/2, we know that the function is negative. But right at 9/2, we can assume that the function is 0, and at x < 9/2 the function is positive. Likewise for x < -1. So we can use 9/2 and -1 as our x-intercepts. So, we have the following:

0 = a(9/2)^2 + b(9/2) + c
0 = (81/4)a + (9/2)b + c

0 = a(-1)^2 + b(-1) + c
0 = -a - b + c

We also have a point at (1, 28) given from the problem. So,

28 = a(1)^2 + b(1) + c
28 = a + b + c

Three equations and three unknowns - we can solve this now. If we added the second and third equations, the a and b terms cancel out, and we can solve for c:

(0 = -a - b + c) + (28 = a + b + c) = (28 = 2c)
c = 14

From here, we can plug into another equation, solve for a (or b), substitute that into the remaining equation, and solve for b (or a). Let's use the second equation first:

0 = -a - b + 14
a = 14 - b

Plug this into the first equation:

0 = (81/4)(14 - b) + (9/2)b + 14
0 = (567/2) - (81/4)b + (9/2)b + 14
0 = (595/2) - (63/4)b
(63/4)b = (595/2)
b = 1190/63
b = 170/9

And finally, we can solve for a:

a = 14 - (170/9)
a = -44/9

The answer for a makes sense, since the parabola points down. The negative values are right of 9/2 and left of -1 - the positive values are between the numbers, which means the vertex is positive as well.

So your equation then is y = -(44/9)x^2 + (170/9)x + 14.