The x-intercepts are the x-values of the points where the graph intercepts the x-axis.
Graph:
x-intercepts: x ≈ -2.16, x=1, and x ≈ 1.75 Local Maximum: (- 1.63,10.47) Local Minimum: (1.46, -1.68) Increasing Interval: x < - 1.63 and x > 1.46 Decreasing Interval: - 1.63 < x < 1.46
Consider the given function.
f(x)= x^5-4x^3+x^2+2
Let's draw the graph. If you need a step-by-step explanation on graphing a function using a table of values, please refer to the extra information at the bottom of this exercise.
x-coordinates of the points at which the graph intercepts the x-axis
The lowest and highest point for a region of the graph.
In the Given Function
x ≈ -2.16, x=1, and x ≈ 1.75
(- 1.63,10.47) and (1.46, -1.68)
Identify x-intercepts and Local Extrema
A local maximum is the point where the function goes from increasing interval to decreasing interval. A local minimum is the point where the function goes from decreasing interval to increasing interval.
As seen on the graph the function has one local maximum and one local minimum.
Local Maximum:& (- 1.63,10.47)
Local Minimum:& (1.46, -1.68)
Determine the Intervals
To determine the intervals for which the function is increasing, decreasing, and constant, let's consider the graph of the function as we move along it in the positive x-direction.
This graph increases for values of x less than - 1.63 and greater than 1.46. It decreases for values of x between - 1.63 and 1.46.
Increasing Interval:& x < - 1.63 and x > 1.46
Decreasing Interval:& - 1.63 < x < 1.46
Showing Our Work
Graphing the function using a table of values
To graph the function, we will use a table to plug in x-values and get the corresponding y-values.
x
x^5-4x^3+x^2+2
f(x)= x^5-4x^3+x^2+2
- 2
( - 2)^5-4( - 2)^3+( - 2)^2+2
6
0
( 0)^5-4( 0)^3+( 0)^2+2
2
1
( - 1)^5-4( - 1)^3+( - 1)^2+2
0
2
( 2)^5-4( 2)^3+( 2)^2+2
6
We will now plot the obtained points in a coordinate plane and connect them with a non-linear line.
