Answer
Step 1: $y=2x^4$.
Step 2: x-intercepts $x=0,-6,\pm2$; y-intercepts $y=0$.
Step 3: $x=0,-6,\pm2$ (all with multiplicity 1, cross the x-axis)..
Step 4: $3$.
Step 5: See graph.
Work Step by Step
Step 1: Given $f(x)=2x^4+12x^3-8x^2-48x=2x^3(x+6)-8x(x+6)=2x(x+6)(x^2-4)=2x(x+6)(x+2)(x-2)$, the end behavior of the graph is similar to $y=2x^4$.
Step 2: To find the x-intercepts, let $f(x)=0$ to get $x=0,-6,\pm2$; to find the y-intercepts, let $x=0$ to get $y=0$.
Step 3: We can find the zeros as $x=0,-6,\pm2$ (all with multiplicity 1, cross the x-axis)..
Step 4: The maximum number of turning points is $n-1=3$.
Step 5: See graph.
