Question

determine if the function shown on the graph is continuous. if not, identify the type (infinite, jump, or removable) and location of discontinuity.
1.
continuous
discontinuous
type: ____________
location: ____________
use the graph to determine if the relation is symmetrical to the x - axis, y - axis, and/or origin.

1. $y = x^{4}-4x^{2}$

x - axis
y - axis
origin
none
describe the end behavior of the graph.
5.

Explanation:

Step1: Check continuity for first - graph

A function is continuous if there are no breaks, jumps, or holes. The first graph has a hole, so it is discontinuous.

Step2: Identify type of discontinuity

A removable discontinuity is a hole in the graph. Since the first graph has a hole, it is a removable discontinuity.

Step3: Locate discontinuity

The hole in the first graph is at the x - value where the open - circle is located. By observing the graph, it is at \(x = 3\).

Step4: Check symmetry for \(y=x^{4}-4x^{2}\)

Replace \(x\) with \(-x\): \(y=(-x)^{4}-4(-x)^{2}=x^{4}-4x^{2}\). Since \(y(-x)=y(x)\), the function is symmetric about the y - axis. If it were symmetric about the x - axis, replacing \(y\) with \(-y\) would give the original equation, which is not the case. If it were symmetric about the origin, replacing \(x\) with \(-x\) and \(y\) with \(-y\) would give the original equation, which is not true here.

Step5: Analyze end - behavior for second graph

As \(x\to+\infty\), the function value \(y\to0\) (the graph approaches the x - axis from above). As \(x\to-\infty\), the function value \(y\to0\) (the graph approaches the x - axis from above).

Answer:

1. Discontinuous

Type: Removable
Location: \(x = 3\)

1. y - axis
2. As \(x\to+\infty\), \(y\to0\); as \(x\to-\infty\), \(y\to0\)