Question

Use the x-intercepts to find the intervals on which the graph of f is above and below the x-axis.

f(x) = (x + 3)²
A) Above: x > -3; Below: x < -3
B) Above: x < -3; Below: x > -3
C) Above: x < -3; Below: x > -3
D) Above: x > -3; Below: x < -3

Answer 1

The graph of f(x) = (x + 3)² is above the x-axis for all values of x, as it is a parabola that opens upwards and only touches the x-axis at its vertex (x = -3). There are no intervals where the graph is below the x-axis.

Step-by-step explanation:

To determine the intervals on which the graph of f(x) = (x + 3)² is above or below the x-axis, we need to consider its x-intercepts and the nature of the function. Since the function is a perfect square, it only touches the x-axis at its vertex, which is at x = -3. Furthermore, since the parabola opens upwards (the coefficient of x² is positive), the graph will be above the x-axis for all other values of x.

Therefore, the graph of f(x) is above the x-axis for all x > -3 and also for x < -3, meaning it is never below the x-axis. The correct answer is thus:

• Above: x > -3 and x < -3

There is no interval where the graph is below the x-axis since the function represents a parabola that does not cross the x-axis. As such, options B, C, and D are incorrect because they suggest intervals where the parabola is below the x-axis, which does not happen for this function.