To determine which function represents the described parabola, let's analyze the key characteristics of the parabola based on the information given:
-
Opens Up: This means the coefficient of the squared term is positive. All functions listed have a positive coefficient since they are in the form of \((x - h)^2 + k\).
-
Vertex in Quadrant 4: The y-coordinate of the vertex \(k\) must be negative (for the vertex to be below the x-axis) and the x-coordinate \(h\) can be positive (quadrant 4).
-
Crosses the x-axis into Quadrant 1: The parabola must start in quadrant 1, cross the x-axis, and return to quadrant 1, which means it has two x-intercepts.
Let's analyze each function:
-
\(f(x) = (x – 8)^2 + 6\):
- Vertex: \( (8, 6) \) - Not in quadrant 4, it's in quadrant 1.
-
\(f(x) = (x + 8)^2 + 6\):
- Vertex: \( (-8, 6) \) - Not in quadrant 4; it's in quadrant 2.
-
\(f(x) = (x + 8)^2 - 6\):
- Vertex: \( (-8, -6) \) - This is in quadrant 4 since the y-coordinate is negative, and the x-coordinate is negative.
-
\(f(x) = (x – 8)^2 - 6\):
- Vertex: \( (8, -6) \) - This is also in quadrant 4, with a positive x-coordinate and negative y-coordinate.
Now, let's check the x-intercepts:
-
For \(f(x) = (x + 8)^2 - 6\): \[ -6 = 0 \implies (x + 8)^2 = 6 \implies x + 8 = \pm \sqrt{6} \] The solutions are: \[ x = -8 + \sqrt{6}, \quad x = -8 - \sqrt{6} \] One of these is in quadrant 1 since \(-8 + \sqrt{6}\) is slightly less than -8 but greater than -9.
-
For \(f(x) = (x - 8)^2 - 6\): \[ -6 = 0 \implies (x - 8)^2 = 6 \implies x - 8 = \pm \sqrt{6} \] The solutions are: \[ x = 8 + \sqrt{6}, \quad x = 8 - \sqrt{6} \] Both of these will cross into quadrant 1.
From this analysis, \(f(x) = (x + 8)^2 - 6\) has its vertex in quadrant 4 and crosses the x-axis, going into quadrant 1.
Thus, the answer is:
\(f(x) = (x + 8)^2 - 6\).
1 answer