Identify any intercepts, relative extrema, points of inflection, and asymptotes. (If an answer does not exist, enter DNE.)
y = x^3 + 3x^2 + 3x + 2
x-intercept (x, y) =
y-intercept (x, y) =
relative maximum (x, y) =
relative minimum (x, y) =
point of inflection (x, y) =
horizontal asymptote
vertical asymptote
To find the intercepts, relative extrema, points of inflection, and asymptotes of the function y = x^3 + 3x^2 + 3x + 2, we'll need to analyze the function and its derivatives.
First, let's find the x-intercepts by setting y = 0 and solving for x:
0 = x^3 + 3x^2 + 3x + 2
There is no simple way to factor or solve this equation, so we'll use numerical methods or a graphing calculator to approximate the x-intercepts. By graphing the function, we can see that there is only one x-intercept at approximately (-2.46, 0).
Next, let's find the y-intercept by setting x = 0:
y = (0)^3 + 3(0)^2 + 3(0) + 2
y = 2
So the y-intercept is (0, 2).
To find the relative extrema, we'll need to find the critical points of the function. These occur where the derivative is zero or undefined. Taking the derivative of the function:
y' = 3x^2 + 6x + 3
Setting y' = 0, we have:
0 = 3x^2 + 6x + 3
0 = x^2 + 2x + 1
0 = (x + 1)^2
This gives us a double root at x = -1. Therefore, the relative minimum occurs at (-1, -2).
Now, let's find the points of inflection. We'll find the second derivative of the function:
y'' = 6x + 6
Setting y'' = 0, we have:
0 = 6x + 6
-6 = 6x
x = -1
So the point of inflection occurs at (-1, -2).
Lastly, let's determine the asymptotes.
There are no horizontal asymptotes because as x approaches positive or negative infinity, the function also approaches positive or negative infinity.
There are also no vertical asymptotes because the function is a polynomial and does not have any vertical asymptotes.
To summarize:
x-intercept: (-2.46, 0)
y-intercept: (0, 2)
relative maximum: does not exist
relative minimum: (-1, -2)
point of inflection: (-1, -2)
horizontal asymptote: none
vertical asymptote: none
To identify intercepts, relative extrema, points of inflection, and asymptotes of the given function, we can follow these steps:
1. Intercepts:
a) x-intercept: Set y = 0 and solve for x.
Substitute y = 0 in the equation: x^3 + 3x^2 + 3x + 2 = 0
We can solve this equation using factoring or synthetic division to find the x-intercepts.
b) y-intercept: Set x = 0 and solve for y.
Substitute x = 0 in the equation: y = 0^3 + 3(0^2) + 3(0) + 2
Compute the value to find the y-intercept.
2. Relative extrema:
To find relative extrema, we need to find the critical points of the function. These points occur where the derivative of the function is equal to zero or undefined.
a) Calculate the first derivative of the function.
b) Set the derivative equal to zero and solve for x.
c) Evaluate the function at these x-values to find the corresponding y-values.
3. Points of inflection:
To find points of inflection, we need to locate the x-values where the concavity of the function changes.
a) Calculate the second derivative of the function.
b) Set the second derivative equal to zero and solve for x.
c) Determine the x-values where the concavity changes by analyzing the sign changes of the second derivative around these x-values.
4. Asymptotes:
a) Horizontal asymptote:
As x approaches positive or negative infinity, check the behavior of the function to determine if there is a horizontal asymptote.
If the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0.
If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients of both polynomials to find the horizontal asymptote.
b) Vertical asymptotes:
Set the denominator equal to zero and solve for x.
These x-values will give the vertical asymptotes of the function.
Following these steps, you can find the x-intercept, y-intercept, relative extrema, points of inflection, horizontal asymptote, and vertical asymptotes for the given function y = x^3 + 3x^2 + 3x + 2.