Let's start by finding the minimum point of the quadratic graph y = 3x² - ax + b. The minimum point of a quadratic function in the form y = ax² + bx + c occurs at the vertex, and its x-coordinate is given by:
x_vertex = -b / (2a)
In this case, a = 3, so:
x_vertex = -(-a) / (2 * 3) = a / 6
Now, we're given that this minimum point lies on the linear function y = (-1/2)x. So, we can set the y-values of these two functions equal to each other:
3(a/6)² - a(a/6) + b = (-1/2)(a/6)
Simplify this equation:
(a²/12) - (a²/6) + b = (-a/12)
Now, let's isolate b:
b = (-a/12) - (a²/12) + (a²/6)
Combine the terms on the right side:
b = (-a/12) + (a²/12)
Now, we have an expression for b in terms of a:
b = (a² - a) / 12
Next, we're given that the graph passes through the point (4, 9), which means when x = 4, y = 9. We can use this information to form another equation:
3(4)² - a(4) + b = 9
Simplify and substitute the expression for b:
48 - 4a + (a² - a) / 12 = 9
Multiply through by 12 to get rid of the fraction:
12(48) - 48a + a² - a = 12(9)
Now, simplify:
576 - 48a + a² - a = 108
Rearrange and combine like terms:
a² - 49a + 468 = 0
Now, we have a quadratic equation in terms of a. To find the solutions for a, you can use the quadratic formula:
a = (-b ± √(b² - 4ac)) / (2a)
In this case, a = 1, b = -49, and c = 468. Plug these values into the quadratic formula to find the solutions for a.
Please note that these solutions may not necessarily be integers, as the problem does not specify that a and b must be integers.