Answer
a. $x=-\frac{1}{5},1,\pm 2i$
b. See graph and explanations.
Work Step by Step
a. Based on the graph, we can identify a zero as $x=1$. However, as the function is of $4th$ degree, we need to find another zero to reduce it to a quadratic form. Using synthetic division with possible zeros of $\pm1,\pm2,\pm4,\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5}$, we can find an additional zero at $x=-\frac{1}{5}$ and the quotient as $-5x^2-20=-5(x^2+4)$ (shown in the figure). Thus the zeros are $x=-\frac{1}{5},1,\pm 2i$
b. The end behaviors of the function can be found as $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$ . The maximum number of of turning points is $4-1=3$ (reduced to 1 due to imaginal zeros), and the y-intercept can be found as $y=f(0)=4$. With the above information, we can finish the graph as shown in the figure, where the locations and values of the extrema are not a real concern at this point.
