Solving for 'a' and 'b' in a Quadratic Equation Passing Through Two Points
The question in the image provides an equation of a graph, y = ax^2 + bx, and mentions that this graph passes through two points: (1,1) and (3,2). To find the values of 'a' and 'b', we can substitute these points into the equation to form a system of linear equations, which we can then solve to find the values of 'a' and 'b'.
Let's substitute point (1,1):
1 = a(1)^2 + b(1)
1 = a + b ... (1)
Now, we substitute point (3,2):
2 = a(3)^2 + b(3)
2 = 9a + 3b ... (2)
We now have a system of two equations with two variables. Equation (1) can be rewritten as b = 1 - a.
Let's substitute b from equation (1) into equation (2):
2 = 9a + 3(1 - a)
2 = 9a + 3 - 3a
2 = 6a + 3
Now we solve for 'a':
6a = 2 - 3
6a = -1
a = -1/6
With 'a' found, we can now find 'b' by substituting 'a' back into equation (1):
1 = (-1/6) + b
b = 1 + 1/6
b = 6/6 + 1/6
b = 7/6
Therefore, the values of 'a' and 'b' are:
a = -1/6
b = 7/6
