To find the y-value of the vertex of the quadratic function f(x) = x^2 – 3x + 11, we first need to identify the vertex using the standard formula for a quadratic equation in the form of f(x) = ax^2 + bx + c, where:
- a = coefficient of x2
- b = coefficient of x
- c = constant term
In the given function:
- a = 1
- b = -3
- c = 11
The x-coordinate of the vertex can be calculated using the formula:
x = -b / (2a)
Plugging in the values, we get:
x = -(-3) / (2 * 1) = 3 / 2 = 1.5
Next, we need to find the corresponding y-coordinate by substituting the x-value back into the function:
f(1.5) = (1.5)2 – 3(1.5) + 11
This simplifies as follows:
- f(1.5) = 2.25 – 4.5 + 11
- f(1.5) = 8.75
Thus, the y-value of the vertex for the quadratic function f(x) = x^2 – 3x + 11 is 8.75.