The range of a quadratic function is the set of real numbers less than or equal to -1. Which equations could represent that function
Select TWO correct answers
A.) f(x) = 2( x - 2 )^2 - 1
B.) f(x) = -3( x + 1 )^2 - 1
C.) f(x) = -4( x - 1 )^2 +1
D.) f(x) = -( x + 2 )^2 -1
E.) f(x) = ( x + 1 )^2 + 1
Quadratic functions are of the form f(x) = ax^2 + bx + c. If a < 0, the parabola opens downward, and the vertex represents the maximum point of the function. In this case, since the range of the function is all real numbers less than or equal to -1, we know that:
1. The coefficient 'a' must be negative, so that the parabola opens downward.
2. The y-coordinate of the vertex will be -1 because this is the maximum value of the function.
Given these criteria, let's examine the options:
A.) f(x) = 2(x - 2)^2 - 1
Here, 'a' is positive (2), so the graph opens upward. This cannot represent a function with a maximum value of -1. This is incorrect.
B.) f(x) = -3(x + 1)^2 - 1
Here, 'a' is negative (-3), and the vertex form of the quadratic shows that the vertex is at (-1, -1). This meets the criteria for the range. This is one correct option.
C.) f(x) = -4(x - 1)^2 + 1
Here, 'a' is negative (-4), but the y-coordinate of the vertex is +1, which is not the maximum value for the given range. Therefore, it does not satisfy the requirement. This is incorrect.
D.) f(x) = -(x + 2)^2 - 1
Here, 'a' is negative, and the quadratic is translated 2 units to the left and 1 unit down. This puts the vertex at (-2, -1), which means the range would be less than or equal to -1, fitting the requirements. This is another correct option.
E.) f(x) = (x + 1)^2 + 1
Here, 'a' is positive, so the graph opens upward, again not fitting the requirements of the given range. This is incorrect.
The correct answers are B and D.