Answer
a. $x\to-\infty, y\to-\infty$ and when $x\to\infty, y\to-\infty$, see explanations.
b. $x=-4,0,4$, crosses the x-axis at $x=\pm4$, 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 $f(x)=-x^4+16x^2$, we have:
a. The leading term is $-x^4$ with an coefficient of $-1$ and 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 $f(x)=-x^2(x^2-16)=-x^2(x+4)(x-4)$; thus the x-intercepts are $x=-4,0,4$ and the graph crosses the x-axis at the intercept $x=\pm4$, but will touch and turn around at $x=0$ (even powers).
c. We can find the y-intercept by letting $x=0$ which gives $y=0$.
d. Test $f(-x)=-(-x)^4+16(-x)^2=-x^4+16x^2$. 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.
