I need help with this question still!!!
Find the values for a, b, and c such that the graph of y= ax^2 + bx + c has a relative maximum at (3,12) and crosses the y-axis at (0,1).
Please Help!!!
Check your 3-24-11, 2:44pm post.
To find the values of a, b, and c, we need to use the given information about the relative maximum and the y-axis crossing.
1. Relative Maximum at (3,12):
A relative maximum occurs when the slope of the function changes from positive to negative. In a quadratic function, this happens at the vertex of the parabola. The x-coordinate of the vertex is given as 3.
The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
Since the x-coordinate of the vertex is 3, we can set up the equation:
3 = -b / (2a)
2. Crosses the y-axis at (0,1):
When a function crosses the y-axis, the x-coordinate is always 0. We can substitute x = 0 and y = 1 into the equation y = ax^2 + bx + c:
1 = a(0)^2 + b(0) + c
1 = c
Now we have two equations:
3 = -b / (2a)
1 = c
Let's solve these equations to find the values of a, b, and c.
From equation 1:
3 = -b / (2a)
Cross multiplying, we get:
6a = -b
Substituting this into equation 2:
1 = c
So, we have found that c = 1.
Substituting the value of c back into equation 1:
6a = -b
Rearranging this equation, we get:
b = -6a
Now, we have the values of c and b. Let's find a.
Substituting the values of b and c into the original equation:
y = ax^2 + bx + c
12 = a(3)^2 + (-6a)(3) + 1
12 = 9a - 18a + 1
12 = -9a + 1
11 = -9a
a = -11 / 9
Therefore, the values for a, b, and c that satisfy the given conditions are:
a = -11 / 9
b = -6a = -6(-11 / 9) = 22 / 3
c = 1