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{{ printedBook.courseTrack.name }} {{ printedBook.name }} Let's begin by determining the degree of each function. It is determined by the exponent with the highest value. The polynomial function $p(x)=0.6x_{3}+2x_{2}+4$ is of degree $3,$ $q(x)=x+4$ is of degree $1$ and $h(x)=0.2x_{2}+x+4$ is of degree $2.$ Now we need to link each graph to its corresponding function by thinking about what the functions' graphs *should* look in order for them to match.

Graph A has the following end behavior. $Asx→-∞,f(x)→-∞andAsx→+∞,f(x)→∞ $ That is consistent with a function with odd degree. We have two polynomial functions with odd degree, $p(x)$ and $q(x).$ Since the graph is not a linear function, then $q(x)$ is not a match. Therefore, we know that the graph must be that of the function $p(x)=0.6x_{3}+2x_{2}+4.$

Now there are two functions left, $h(x)$ and $q(x).$ Since the graph B is a parabola, the function has a degree of $2.$ Therefore, the function $h(x)=0.2x_{2}+x+4$ is the function for this graph.

Since the blue graph is a line it represents a linear function. Thus, we can conclude that it is the graph of $q(x)=x+4.$