Answer
a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to\infty$, see explanations.
b. $x=-1,0,1$, crosses the x-axis at $x=\pm1$, touch and turn around at $x=0$
c. $y=0$.
d. symmetric with respect to the y-axis .
e. See graph and explanations.
Work Step by Step
Given the function $f(x)=x^4-x^2$, we have:
a. The leading term is $x^4$ with an even power, when $x\to-\infty, y\to\infty$ and when $x\to\infty, y\to\infty$
Thus the curve will rise as $x$ increases (right end) and it will also rise as $x$ decreases (left end).
b. Factor the equation as $f(x)=x^2(x^2-1)=x^2(x+1)(x-1)$.
Thus the x-intercepts are $x=-1,0,1$ and the graph crosses the x-axis at the intercept $x=\pm1$, but touches and turns around at $x=0$ (even powers).
c. We can find the y-intercept by letting $x=0$ which gives $y=0$.
d. Test $f(-x)=(-x)^4-(-x)^2=x^4-x^2$. As $f(-x)= f(x)$, the graph is symmetric with respect to the y-axis.
e. See graph; as $n=4$, the maximum number of turning points will be $3$ which agrees with the graph.
