Answer
a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$
b. $x=\pm1$, crosses the x-axis
c. $y=\frac{1}{2}$.
d. symmetric with respect to the y-axis.
e. See graph
Work Step by Step
Given the function $f(x)=-\frac{1}{2}x^4+\frac{1}{2}$, we have:
a. The leading term is $-\frac{1}{2}x^4$ with a coefficient of $-\frac{1}{2}$ and an even power; thus $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$, and the end behaviors are that the curve will fall as $x$ increases (right end) and it will also fall as $x$ decreases (left end).
b. Factor the equation as $f(x)=-\frac{1}{2}(x^4-1)=-\frac{1}{2}(x^2+1)(x+1)(x-1)$; thus the x-intercepts are $x=\pm1$ and the graph crosses the x-axis at $x=\pm1$ (odd multiplicity),
c. We can find the y-intercept by letting $x=0$, which gives $y=\frac{1}{2}$.
d. Test $f(-x)=-\frac{1}{2}(-x)^4+\frac{1}{2}=-\frac{1}{2}x^4+\frac{1}{2}$. 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.
