Answer
a. $x=\frac{1}{2},3,-1\pm i$
b. See graph and explanations.
Work Step by Step
a. Based on the graph, we can identify a zero as $x=\frac{1}{2}$. 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 $\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}$, we can find an additional zero at $x=3$ and the quotient as $2x^2+4x+4=2(x^2+2x+2)$, as shown in the figure. Thus the zeros are $x=\frac{1}{2},3,-1\pm i$
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 turning points is $4-1=3$ (reduced to 1 due to imaginal zeros), and the y-intercept can be found as $y=f(0)=6$. 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.
