To find the maximum or minimum value of the quadratic function y = 2x² – 32x + 12, we will first identify the coefficients in the standard form of the quadratic equation y = ax² + bx + c. Here, a = 2, b = -32, and c = 12.
Since the coefficient of x² (which is a = 2) is positive, the parabola opens upwards. This means that the function has a minimum value, not a maximum.
We can find the x-coordinate of the vertex, which gives us the minimum value of the function, using the formula x = -b / (2a).
- b = -32
- a = 2
Plugging in the values:
x = -(-32) / (2 * 2) = 32 / 4 = 8
Now, we can find the minimum value by substituting x = 8 back into the function:
y = 2(8)² – 32(8) + 12
y = 2(64) – 256 + 12
y = 128 – 256 + 12
y = -128 + 12 = -116
So, the minimum value of the function is -116.
Next, we’ll determine the range of the function. Since the parabola opens upwards and the minimum value is -116, the range of the function is all real numbers that are greater than or equal to -116.
Thus, the range of the function is:
Range: [ -116, ∞ )