match each polynomial function to its graph.\nf(x)=-9x^{4}\ng(x)=-x^{3}+18x^{2}-108x + 216=-(x…

match each polynomial function to its graph.\nf(x)=-9x^{4}\ng(x)=-x^{3}+18x^{2}-108x + 216=-(x - 6)^{3}\nh(x)=-x^{4}-4x^{2}-3=-(x^{2}+3)(x^{2}+1)\nk(x)=-x^{3}+20x^{2}-132x + 288=-(x - 6)^{2}(x - 8)\nf(x)=-9x^{4}\ng(x)=-x^{3}+18x^{2}-108x + 216\nh(x)=-x^{4}-4x^{2}-3\nk(x)=-x^{3}+20x^{2}-132x + 288
Answer
Explanation:
Step1: Analyze the end - behavior of $f(x)=-9x^{4}$
Since the degree $n = 4$ (even) and the leading coefficient $a=-9<0$, as $x\to\pm\infty$, $y\to-\infty$.
Step2: Analyze the end - behavior of $g(x)=-(x - 6)^{3}$
The degree $n = 3$ (odd) and the leading coefficient $a=-1<0$. As $x\to-\infty$, $y\to\infty$ and as $x\to\infty$, $y\to-\infty$. Also, it has a triple root at $x = 6$.
Step3: Analyze the end - behavior of $h(x)=-(x^{2}+3)(x^{2}+1)$
The degree $n = 4$ (even) and the leading coefficient $a=-1<0$. Since $x^{2}+3>0$ and $x^{2}+1>0$ for all real $x$, $h(x)<0$ for all real $x$. As $x\to\pm\infty$, $y\to-\infty$.
Step4: Analyze the end - behavior of $k(x)=-(x - 6)^{2}(x - 8)$
The degree $n = 3$ (odd) and the leading coefficient $a=-1<0$. It has a double root at $x = 6$ and a single root at $x = 8$. As $x\to-\infty$, $y\to\infty$ and as $x\to\infty$, $y\to-\infty$.
Without the actual graphs shown completely, we assume we can match based on end - behavior and roots: We need to match these functions to their respective graphs by looking at end - behavior (as $x\to\pm\infty$) and the number and nature of roots (where the function crosses or touches the $x$ - axis). But since the graphs are not fully labeled, we can only provide the general approach. If we had the full - labeled graphs, we would match as follows:
- $f(x)=-9x^{4}$: Even - degree, negative leading coefficient, graph opens down and is symmetric about the $y$ - axis.
- $g(x)=-(x - 6)^{3}$: Odd - degree, negative leading coefficient, graph passes through $x = 6$ with a horizontal tangent at that point (due to triple root).
- $h(x)=-(x^{2}+3)(x^{2}+1)$: Even - degree, negative leading coefficient, no real roots (since $x^{2}+3>0$ and $x^{2}+1>0$ for all real $x$), graph is always below the $x$ - axis.
- $k(x)=-(x - 6)^{2}(x - 8)$: Odd - degree, negative leading coefficient, touches the $x$ - axis at $x = 6$ (double root) and crosses at $x = 8$.
Answer:
Without the actual labeled graphs, we cannot provide a specific one - to - one match. But the above analysis gives the characteristics to use for matching.