The function in question is a quadratic function expressed as:
y = 2x² – 32x + 12
This function opens upwards since the coefficient of the x² term (2) is positive. To find the maximum or minimum value of this function, we need to determine the vertex of the parabola.
The x-coordinate of the vertex of a parabola given by the general form y = ax² + bx + c can be found using the formula:
x = -b/(2a)
In our case, a = 2 and b = -32. Plugging these values into the formula gives:
x = -(-32)/(2 * 2) = 32/4 = 8
Now to find the corresponding y-value (the minimum value of the function) at this x-coordinate, we substitute x = 8 back into the original function:
y = 2(8)² – 32(8) + 12
y = 2(64) – 256 + 12
y = 128 – 256 + 12
y = -116
Therefore, the minimum value of the function y = 2x² – 32x + 12 is -116 at x = 8.
Since the function opens upward, there is no maximum value (the function approaches infinity as x moves away from the vertex). Thus, we can conclude:
Maximum Value: None (infinity)
Minimum Value: -116 (at x = 8)
Now, to find the range of the function, we know that the parabola reaches its lowest point at y = -116. Because it opens upwards, the function can take on any value greater than or equal to this minimum value.
Thus, the range of the function is:
Range: y ≥ -116
To summarize:
- Minimum Value: -116 (at x = 8)
- Maximum Value: None
- Range: y ≥ -116