What is the maximum or minimum value of the function? What is the range? y = –2x2 + 32x –12 :
116 range: y 116 maximum:
116 range: 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 need to determine the vertex of the quadratic equation. The vertex can be found using the formula x = -b/2a.
In this case, a = -2 and b = 32.
x = -32/(2*(-2))
x = -32/(-4)
x = 8
To find the y-value of the vertex, we substitute the value of x into the equation:
y = -2(8)^2 + 32(8) - 12
y = -2(64) + 256 - 12
y = -128 + 256 - 12
y = 116
Therefore, the vertex is (8, 116).
Since the coefficient of the x^2 term is negative (-2), the parabola opens downwards and the vertex represents the maximum value of the function. Therefore, the maximum value of the function is 116.
The range of the function is all the possible values of y. Since the parabola opens downwards and the maximum value is 116, the range will be all real numbers less than or equal to 116.
To find the maximum or minimum value of the function y = -2x^2 + 32x - 12, we can consider its quadratic form which is in the form of ax^2 + bx + c.
The coefficient of the x^2 term is -2, which means that the parabola opens downward. Hence, the function has a maximum value.
To find the x-coordinate of the vertex, we can use the formula x = -b/2a. In this case, a = -2 and b = 32. Plugging these values into the formula, we get x = -32 / (2*(-2)) = -32 / -4 = 8.
To find the y-coordinate of the vertex, we can substitute the x-coordinate (x = 8) into the function:
y = -2(8)^2 + 32(8) - 12
= -2(64) + 256 - 12
= -128 + 256 - 12
= 116
So, the maximum value of the function is 116.
Now let's consider the range of the function. Since the parabola opens downward, the y-values decrease as the x-values increase. Therefore, the range of the function is all real numbers less than or equal to 116.
In summary:
- The maximum value of the function is 116.
- The range of the function is y ≤ 116.
To find the maximum or minimum value of a function, you need to determine whether it is a maximum or minimum point. In this case, the given function is a quadratic function in the form of y = -2x^2 + 32x - 12.
To find the maximum or minimum point of a quadratic function, you can use the formula x = -b/(2a), where a, b, and c are coefficients of the quadratic equation in the form of ax^2 + bx + c.
In the given function, a = -2 and b = 32. Plugging these values into the formula, we get x = -32/(2*-2) = -32/(-4) = 8.
To find the corresponding y-value, substitute the x-value into the given function:
y = -2(8)^2 + 32(8) - 12
y = -128 + 256 - 12
y = 116
Therefore, the maximum (or minimum) value of the function is 116.
The range of a function refers to all possible output values, or the y-values. Since the maximum value is 116, and there are no restrictions mentioned in the problem, the range of the given function is (negative infinity, 116].
In summary:
- The maximum value of the function is 116.
- The range of the function is (negative infinity, 116].