graph the function y = 2x^4 + 4x^3 = 2x^3(x + 2) by identifying the domain and any symmetries, finding the…

Answer:

1. Domain:
   • The function (y = 2x^{4}+4x^{3}=2x^{3}(x + 2)) is a polynomial function. The domain of a polynomial function is all real - numbers, i.e., ((-\infty,\infty)).
2. Symmetries:
   • Replace (x) with (-x): (y=2(-x)^{4}+4(-x)^{3}=2x^{4}-4x^{3}). Since (y(-x)\neq y(x)) and (y(-x)\neq - y(x)), the function has no symmetry about the (y) - axis or the origin.
3. First - derivative:
   • Using the power rule ((x^{n})^\prime=nx^{n - 1}), if (y = 2x^{4}+4x^{3}), then (y^\prime=8x^{3}+12x^{2}=4x^{2}(2x + 3)).
4. Critical points:
   • Set (y^\prime = 0), so (4x^{2}(2x + 3)=0).
     • (4x^{2}=0) gives (x = 0) and (2x+3 = 0) gives (x=-\frac{3}{2}).
5. Increasing and decreasing intervals:
   • Consider the intervals ((-\infty,-\frac{3}{2})), ((-\frac{3}{2},0)), and ((0,\infty)).
     • For (x\in(-\infty,-\frac{3}{2})), let (x=-2). Then (y^\prime=4(-2)^{2}[2(-2)+3]=16(-1)=-16\lt0), so the function is decreasing on ((-\infty,-\frac{3}{2})).
     • For (x\in(-\frac{3}{2},0)), let (x =-\frac{1}{2}). Then (y^\prime=4(-\frac{1}{2})^{2}[2(-\frac{1}{2})+3]=1\times2 = 2\gt0), so the function is increasing on ((-\frac{3}{2},0)).
     • For (x\in(0,\infty)), let (x = 1). Then (y^\prime=4(1)^{2}(2\times1 + 3)=4\times5 = 20\gt0), so the function is increasing on ((0,\infty)).
6. Second - derivative:
   • Differentiate (y^\prime=8x^{3}+12x^{2}) with respect to (x). Using the power rule, (y^{\prime\prime}=24x^{2}+24x=24x(x + 1)).
7. Inflection points:
   • Set (y^{\prime\prime}=0), so (24x(x + 1)=0). This gives (x = 0) and (x=-1).
8. Concavity:
   • Consider the intervals ((-\infty,-1)), ((-1,0)), and ((0,\infty)).
     • For (x\in(-\infty,-1)), let (x=-2). Then (y^{\prime\prime}=24(-2)(-2 + 1)=48\gt0), so the function is concave up on ((-\infty,-1)).
     • For (x\in(-1,0)), let (x=-\frac{1}{2}). Then (y^{\prime\prime}=24(-\frac{1}{2})(-\frac{1}{2}+1)=-6\lt0), so the function is concave down on ((-1,0)).
     • For (x\in(0,\infty)), let (x = 1). Then (y^{\prime\prime}=24(1)(1 + 1)=48\gt0), so the function is concave up on ((0,\infty)).
9. Asymptotes:
   • Since it is a polynomial function, there are no vertical or horizontal asymptotes.
10. Intercepts:

• x - intercepts: Set (y = 0), so (2x^{3}(x + 2)=0). This gives (x = 0) and (x=-2).
• y - intercept: Set (x = 0), then (y = 0).

11. Extreme points:

• Since the function changes from decreasing to increasing at (x=-\frac{3}{2}), (y(-\frac{3}{2})=2(-\frac{3}{2})^{4}+4(-\frac{3}{2})^{3}=2\times\frac{81}{16}-4\times\frac{27}{8}=\frac{81}{8}-\frac{108}{8}=-\frac{27}{8}).
• At (x = 0), the function does not change from increasing to decreasing or vice - versa (the derivative has a double root at (x = 0)), so it is not a local extreme.
• As (x\to\pm\infty), (y\to\infty) since the leading term (2x^{4}) has a positive leading coefficient and an even degree. So the absolute minimum is ((-\frac{3}{2},-\frac{27}{8})) and there is no absolute maximum.

To choose the correct graph, we use the information about intercepts ((x=-2,0) and (y = 0)), the behavior at critical points ((x =-\frac{3}{2}) is a local minimum), and concavity.

Explanation:

Step1: Find the domain

Polynomial, domain is ((-\infty,\infty)).

Step2: Check for symmetries

Substitute (-x), no symmetry.

Step3: Calculate the first - derivative

Use power rule, (y^\prime=8x^{3}+12x^{2}).

Step4: Find critical points

Set (y^\prime = 0), solve (4x^{2}(2x + 3)=0).

Step5: Determine increasing and decreasing intervals

Test points in sub - intervals of the domain.

Step6: Calculate the second - derivative

Differentiate (y^\prime), (y^{\prime\prime}=24x^{2}+24x).

Step7: Find inflection points

Set (y^{\prime\prime}=0), solve (24x(x + 1)=0).

Step8: Determine concavity

Test points in sub - intervals for (y^{\prime\prime}).

Step9: Check for asymptotes

No asymptotes for polynomial.

Step10: Find intercepts

Set (x = 0) for y - intercept, (y = 0) for x - intercepts.

Step11: Find extreme points

Analyze behavior at critical points.