Question

which quadratic function has a range that can be described as y < 10.5?
a f(x) = -6(x+4)^2 + 10.5
b f(x) = 6(x+4)^2 + 10.5
c f(x) = -6(x - 10.5)^2 + 4
d f(x) = 10.5(x - 4)^2 - 6

Answer 1

To determine which quadratic function has a range described by \( y < 10.5 \), we need to analyze the vertex form of each function. The general form of a quadratic function in vertex form is:

\[ f(x) = a(x - h)^2 + k, \]

where \( (h, k) \) is the vertex of the parabola, and the sign of \( a \) determines whether the parabola opens upwards (if \( a > 0 \)) or downwards (if \( a < 0 \)).

For the range to be \( y < 10.5 \), the parabola must open downwards and have a maximum value of 10.5. In other words, the vertex \( k \) must equal 10.5 and \( a \) must be negative.

Now let's analyze each function:

a) \( f(x) = -6(x+4)^2 + 10.5 \)
- Vertex: \( (−4, 10.5) \)
- \( a = -6 < 0 \)
- Range: \( y < 10.5 \)

b) \( f(x) = 6(x+4)^2 + 10.5 \)
- Vertex: \( (−4, 10.5) \)
- \( a = 6 > 0 \)
- Range: \( y \geq 10.5 \)

c) \( f(x) = -6(x-10.5)^2 + 4 \)
- Vertex: \( (10.5, 4) \)
- \( a = -6 < 0 \)
- Range: \( y < 4 \)

d) \( f(x) = 10.5(x-4)^2 - 6 \)
- Vertex: \( (4, -6) \)
- \( a = 10.5 > 0 \)
- Range: \( y \geq -6 \)

From the analysis, the only function that has a range of \( y < 10.5 \) is:

**a)** \( f(x) = -6(x+4)^2 + 10.5 \)