What is the maximum or minimum value of the function? What is the range? y = –2x2 + 32x –12 (1 point) Responses maximum: 116 range: y 116 maximum: 116 range: y Image with alt text: A symbol less than or equal to 116 maximum: –116 range: y –116 maximum: –116 range: y Image with alt text: greater than equal to –116 maximum: 116 range: y 116 maximum: 116 range: y Image with alt text: greater than equal to 116 maximum: –116 range: y –11
To find the maximum or minimum value of the function, we can use the formula -b/2a for the x-coordinate of the vertex of a quadratic function in the form ax^2 + bx + c.
For y = -2x^2 + 32x - 12, the coefficient of x^2 is -2, the coefficient of x is 32, and the constant term is -12.
By substituting these values into the formula -b/2a, we get:
x = -32 / (2 * -2)
x = -32 / -4
x = 8
To find the y-coordinate of the vertex, we substitute the x-coordinate into the function:
y = -2(8)^2 + 32(8) - 12
y = -2(64) + 256 - 12
y = -128 + 256 - 12
y = 116
The maximum value of the function is therefore 116.
To find the range, we consider that a quadratic function with a negative coefficient of x^2 opens downwards and has a vertex at the maximum value.
So, the range of y = -2x^2 + 32x - 12 is y ≤ 116.
To find the maximum or minimum value of the function and the range, we first need to determine whether the parabola opens upward or downward. The given function is in the form y = -2x^2 + 32x - 12, which is a quadratic function.
Since the coefficient of the x^2 term is negative, the parabola opens downward, indicating that it has a maximum value.
To find the x-coordinate of the vertex, we can use the formula x = -b / (2a), where a = -2 and b = 32.
x = -32 / (2 * -2)
x = 8
The x-coordinate of the vertex is 8. To find the y-coordinate, we substitute x = 8 into the function:
y = -2(8)^2 + 32(8) - 12
y = -2(64) + 256 - 12
y = -128 + 256 - 12
y = 116
So, the maximum value of the function is 116.
The range is the set of all possible y-values for the function. Since the parabola opens downward, the range will be from negative infinity up to the y-coordinate of the vertex (116 in this case).
Therefore, the range is y ≤ 116.
To find the maximum or minimum value of a quadratic function, we can use the vertex formula. For a quadratic function in the form of y = ax^2 + bx + c, the x-coordinate of the vertex can be found using the formula x = -b/2a.
In this case, we have the function y = -2x^2 + 32x - 12. We can see that a = -2 and b = 32. By substituting these values into the vertex formula, we can find the x-coordinate of the vertex.
x = -(32) / (2 * (-2))
x = -32 / (-4)
x = 8
To find the maximum or minimum value of the function, we substitute the x-coordinate of the vertex 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 maximum value of the function is 116.
Now let's find the range of the function. The range is the set of all possible y-values of the function. In this case, since the maximum value is 116, and the coefficient of the x^2 term is negative (-2), we can determine that the range is all real numbers less than or equal to 116.
Therefore, the range is y ≤ 116.