Answer
a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to\infty$, see explanations.
b. $x=0,1$, touch and turn around at $x=0,1$
c. $y=0$.
d. neither
e. See graph and explanations.
Work Step by Step
Given the function $f(x)=x^4-2x^3+x^2$, we have:
a. The leading term is $x^4$ with a coefficient of $+1$ and an 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 $f(x)=x^2(x^2-2x+1)=x^2(x-1)^2$; thus the x-intercepts are $x=0,1$ and the graph will touch and turn around at $x=0,1$ (even multiplicity).
c. We can find the y-intercept by letting $x=0$ which gives $y=0$.
d. Test $f(-x)=(-x)^4-2(-x)^3+(-x)^2=x^4+2x^3+x^2$. 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.
