Answer
a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$, see explanations.
b. $x=-2,0,2$, crosses the x-axis at $x=\pm2$, touches and turns 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 $f(x)=-x^4+4x^2$, 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 curve will fall as $x$ increases (right end) and it will also fall as $x$ decreases (left end).
b. Factor the equation as $f(x)=-x^2(x^2-4)=-x^2(x+2)(x-2)$; thus the x-intercepts are $x=-2,0,2$ and the graph crosses the x-axis at the intercept $x=\pm2$, but will touch and turn around at $x=0$ (even multiplicity).
c. We can find the y-intercept by letting $x=0$, which gives $y=0$.
d. Test $f(-x)=-(-x)^4+4(-x)^2=-x^4+4x^2$. 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.
