Determined the vertex form and the maximum value of the function.
f(x) =x^2 + x^2 +12x+38
First, we need to rearrange the equation into the vertex form: f(x) = a(x-h)^2 + k
To do this, we complete the square for the x^2 terms.
f(x) = x^2 + x^2 +12x + 38
Combine like terms: f(x) = 2x^2 + 12x + 38
Factor out the common factor of 2 from the x^2 and x terms: f(x) = 2(x^2 + 6x) + 38
To complete the square, we take half of the coefficient of the x term, square it, and add it inside the parentheses. In this case, half of 6 is 3, and 3^2 is 9.
f(x) = 2(x^2 + 6x + 9) + 38 - 2(9)
Simplify inside the parentheses: f(x) = 2(x + 3)^2 + 38 - 18
Combine like terms: f(x) = 2(x + 3)^2 + 20
So the vertex form of the function is f(x) = 2(x + 3)^2 + 20.
Since the coefficient of the (x + 3)^2 term is positive, the parabola opens upward, and the vertex represents the minimum point. Therefore, the maximum value of the function is the y-coordinate of the vertex.
The x-coordinate of the vertex is -3, so we substitute this value into the function: f(-3) = 2(-3 + 3)^2 + 20
Simplify: f(-3) = 2(0)^2 + 20 = 20
Therefore, the vertex form of the function is f(x) = 2(x + 3)^2 + 20, and the maximum value of the function is 20.