198,349 views
32 votes
32 votes
Points 1 to 6To present y=-x^3+3x^2, 9 points must be selected, point 1: Domain, 2: zeros of the function, 3: period and symmetry, 4: sign of the function, 5: going to the edge of the function, 6: asymptotes, 7: monotony and extreme values, 8: concavity and convexity, 9: to present the function graphically.

User Kaza
by
3.0k points

1 Answer

17 votes
17 votes

ANSWERS

1. Domain: all real values

2. Zeros: 0 (with multiplicity 2) and 3

3. Not periodic. Symmetric about the point (1, 2)

4. Positive for x < 3; negative for x > 3

5. f → ∞ as x → -∞; f → -∞ as x → ∞

6. none

Step-by-step explanation

1. The domain of a function is the set of all the x-values for which the function exists. In this case, we have a polynomial function and, therefore, the domain is all real values.

2. To find the zeros of the function, we have to solve,


-x^3+3x^2=0

First, factor x² and -1 out. To do so, we have to divide each term by x² and by -1 - or, in other words, divide by -x²,


\begin{gathered} -x^2\left((-x^3)/(-x^2)+(3x^2)/(-x^2)\right)=0 \\ \\ -x^2(x^(3-2)-3x^(2-2))=0 \end{gathered}

So, we have,


-x^2(x-3)=0

In this equation, we can see that if x = 0, then the equation is true. Also, if x = 3 the equation is true. So, these are the two zeros, with the particularity that x = 0 has multiplicity 2. This is because the factor related to that zero is x squared.

Hence, the zeros are 0 and 3. 0 has multiplicity 2.

3. As mentioned before, this is a polynomial function, which means that it is not a periodic function. A cubic function is an odd function, and it is symmetric about the origin. However, this function is not the parent function, x³, but it is symmetric about the point (1, 2).

4. We know that the function is zero at x = 0 and at x = 3. For x < 0, the function is positive,


with\text{ }x=-1:\text{ }y=-(-1)^3+3(-1)^2=-(-1)+3\cdot1=1+3=4

For 0 < x < 3, the function is also positive. This is because x = 0 with multiplicity 2.

Then, since the function crosses the x-axis at x = 3 and that zero has multiplicity 1, we can conclude that the function is negative for x > 3.

Hence, is the function is positive for x < 3 and negative for x > 3.

5. As mentioned in part 4, the function is positive for all values of x less than 3, which means that the function goes to infinity as x goes to negative infinity.

Since for x > 3 the function is always negative, it goes to negative infinity as x goes to infinity.

6. A polynomial function has no restrictions in the domain and, therefore, has no asymptotes.

User Skovorodkin
by
2.5k points