Answer
a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to\infty$,
b. $x=0,3$, touch and turn around
c. $y=0$.
d. neither
e. See graph
Work Step by Step
Given the function $f(x)=x^4-6x^3+9x^2$, we have:
a. The leading term is $x^4$ with a coefficient of $+1$ and even power, when $x\to-\infty, y\to\infty$ and when $x\to\infty, y\to\infty$; thus the end behaviors are that the curve will rise as $x$ increases (right end) and it will also rise as $x$ decreases (left end).
b. Factor the equation as $f(x)=x^2(x^2-6x+9)=x^2(x-3)^2$; thus the x-intercepts are $x=0,3$ and the graph will touch and turn around at $x=0,3$ (even multiplicity).
c. We can find the y-intercept by letting $x=0$ which gives $y=0$.
d. Test $f(-x)=(-x)^4-6(-x)^3+9(-x)^2=x^4+6x^3+9x^2$. As $f(-x)\ne f(x)$ and $f(-x)\ne -f(x)$, the graph is neither symmetric with respect to the y-axis nor with the origin.
e. See graph; as $n=4$, the maximum number of turning points will be $3$, which agrees with the graph.
