Question

which of the following graphs could represent a quartic function?
a. graph a: a red curve with a horizontal and vertical axis
b. graph b: a red line with a horizontal and vertical axis
c. graph c: a red curve with a horizontal and vertical axis
d. graph d: a red vertical line with a horizontal and vertical axis
options: graph a, graph d, graph c (selected), graph b

Explanation:

Step1: Recall the degree - shape relationship

A quartic function is a polynomial function of degree \(n = 4\). The general form of a polynomial function is \(y=a_nx^n+\cdots+a_1x + a_0\), and the end - behavior of a polynomial function is determined by the leading term \(y = a_nx^n\). For a quartic function (\(n = 4\), even degree), if \(a_4>0\), as \(x\to\pm\infty\), \(y\to+\infty\); if \(a_4 < 0\), as \(x\to\pm\infty\), \(y\to-\infty\). Also, the maximum number of turning points of a polynomial function of degree \(n\) is \(n - 1\). For a quartic function (\(n = 4\)), the maximum number of turning points is \(n-1=3\).

Step2: Analyze each graph

• Graph A:
• The end - behavior: as \(x\to-\infty\), \(y\to-\infty\) and as \(x\to+\infty\), \(y\to+\infty\). For a polynomial function, when the degrees are even, the end - behaviors should be the same (both \(+\infty\) or both \(-\infty\)). So, Graph A is not a quartic function.
• Graph B:
• The graph is a straight line, which is a linear function (\(y=mx + b\), degree \(n = 1\)). So, Graph B is not a quartic function.
• Graph C:
• The end - behavior: as \(x\to\pm\infty\), the \(y\) - values have the same direction (if we assume a proper polynomial form). And the number of turning points is \(3\) (which is \(4 - 1\) for a quartic function).
• Graph D:
• The graph is a vertical line, which is not a function (fails the vertical line test: for a function \(y = f(x)\), for each \(x\) in the domain, there is exactly one \(y\) - value. A vertical line \(x = k\) has infinitely many \(y\) - values for \(x=k\)). So, Graph D is not a function.

Answer:

Graph C