Question

fill in the blanks for the attributes of the functions shown in the graphs below. f(x) is the graph 1. f(x) is positive on the interval ______ f(x) has a zero at x = ____ f(x) is increasing on the interval ____ f(x) is decreasing on the interval ____ f(x) has a local minimum of ____ 2. f(x) is positive on the interval ____ f(x) is negative on the interval ____ f(x) has a zero at x = ____ f(x) is increasing on the interval ____ f(x) is decreasing on the interval ______ this is an asymptote

Explanation:

Step1: Analyze the first graph

For the first graph, a parabola opening upwards. The function \(f(x)\) is positive when the graph is above the \(x -\)axis.
The zero is the \(x -\)value where the graph crosses the \(x -\)axis. The function is increasing when the slope is positive and decreasing when the slope is negative. The local - minimum is the lowest \(y -\)value of the function.

1. \(f(x)\) is positive on the interval \((-\infty,- 2)\cup(0,\infty)\) because the graph is above the \(x -\)axis in these intervals.
2. \(f(x)\) has a zero at \(x=-2\) and \(x = 0\) as the graph crosses the \(x -\)axis at these points.
3. \(f(x)\) is increasing on the interval \((-1,\infty)\) as the slope of the graph is positive for \(x>-1\).
4. \(f(x)\) is decreasing on the interval \((-\infty,-1)\) as the slope of the graph is negative for \(x < - 1\).
5. \(f(x)\) has a local minimum of \(-1\) (the \(y -\)value at the vertex \(x=-1\)).

Step2: Analyze the second graph

For the second graph, an exponential - like function with a horizontal asymptote.

1. \(f(x)\) is positive on the interval \((0,\infty)\) as the graph is above the \(x -\)axis for \(x>0\).
2. \(f(x)\) is negative on the interval \((-\infty,0)\) as the graph is below the \(x -\)axis for \(x < 0\).
3. \(f(x)\) has a zero at \(x = 0\) as the graph crosses the \(x -\)axis at \(x = 0\).
4. \(f(x)\) is increasing on the interval \((-\infty,\infty)\) as the slope of the graph is always positive.
5. There is no interval where \(f(x)\) is decreasing since the slope is always non - negative.

Answer:

1.

• \(f(x)\) is positive on the interval \((-\infty,-2)\cup(0,\infty)\)
• \(f(x)\) has a zero at \(x=-2,0\)
• \(f(x)\) is increasing on the interval \((-1,\infty)\)
• \(f(x)\) is decreasing on the interval \((-\infty,-1)\)
• \(f(x)\) has a local minimum of \(-1\)

2.

• \(f(x)\) is positive on the interval \((0,\infty)\)
• \(f(x)\) is negative on the interval \((-\infty,0)\)
• \(f(x)\) has a zero at \(x = 0\)
• \(f(x)\) is increasing on the interval \((-\infty,\infty)\)
• \(f(x)\) is decreasing on the interval None