How do you sketch the curve #y=x^3+6x^2+9x# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

Question

Ezekiel Alexander · Accepted Answer

See graph and explanation

Polynomial graphs have no asymptotes. As x to +-oo, y = x^3(1+3/x)^2 to +-oo, showing end behavior of #uarr and darr#, without limit. So, there are no global extrema. #y=x(x+3)^2=0#, at #x = 0 and -3#. x-intercepts: #0 and -3#. #y'=3(x+1)(x+3)=0#, at #x = -3 and -1# Turning points or points of inflexion at (-1, -4) and (-3, 0) #y''=6x+12=0#, at #x = -2#. #y'''=6 ne 0#. At #(-2, -2), y''=0 and y''' ne 0#. So, it is the point of inflexion. This POI is marked, in the graph. At # (-1, -4)), y'=0 and y''=6> 0#. So, local minimum #y = -4#. At #(-3, 0), y'=0 and y''= -6 #. So,#0# is a local maximum. graph{(x(x+3)^2-y)((x+2)^2+(y+2)^2-.01)=0 [-10, 10, -10, 5]}

Adam Adkins · Answer

undefined To sketch the curve $ y = x^3 + 6x^2 + 9x $, follow these steps:

1. **Find Critical Points:** Set the derivative equal to zero and solve for $ x $ to find critical points.
   
2. **Determine Sign of Derivative:** Use the first derivative test to determine the intervals where the function is increasing or decreasing.
   
3. **Find Inflection Points:** Set the second derivative equal to zero and solve for $ x $ to find inflection points.
   
4. **Determine Concavity:** Use the second derivative test to determine the intervals where the function is concave up or concave down.
   
5. **Find Asymptotes:** Determine any vertical, horizontal, or slant asymptotes by analyzing the behavior of the function as $ x $ approaches infinity or negative infinity.
   
6. **Find Intercepts:** Find the $ y $-intercept by evaluating the function at $ x = 0 $. Find the $ x $-intercepts by setting $ y $ equal to zero and solving for $ x $.

Once you have gathered this information, plot the points, sketch the curve connecting them, and label any relevant features such as maximums, minimums, inflection points, and asymptotes.