How do you use the first and second derivatives to sketch #y= -(x-2) (x+2) (x-4)#?

Question

William Abdalla · Accepted Answer

# y=-(x-2)(x+2)(x-4) #
graph{-(x-2)(x+2)(x-4) [-10, 10, -20, 20]}

Fist we find the roots of the equation, ie the values of #x# st #y=0#: # y=0 => -(x-2)(x+2)(x-4)=0 # # :. (x-2)(x+2)(x-4)=0 # # :. x=2, x=-2, x=4 # Next we find the critical points, ie the values for which #y'=0#, we will need to multiply out as a cubic in order to this: # y=-(x-2)(x+2)(x-4) # # :. y=-(x-2)(x^2-4x+2x-8) # # :. y=-(x-2)(x^2-2x-8) # # :. y=-(x^3-2x^2-8x-2x^2+4x+16) # # :. y=-(x^3-4x^2-4x+16) # # :. y=-x^3+4x^2+4x-16 # So, differentiating wrt #x# gives: # :. y'=-3x^2+8x+4 # # y'=0 => -3x^2+8x+4 = 0# # :. 3x^2-8x-4 = 0# # :. 3x^2-8x-4 = 0 # This quadratic does not factorise, so we use the quadratic formula: # x=(-(-8)+-sqrt((-8)^2-4(3)(-4)))/(2(3)) # # :. x=(8+-sqrt(64+48))/(6) # # :. x=(8+-sqrt(112))/(6) # # :. x=-0.43, 3.10 # (2dp) , or equally we can solve by completing the square: # 3x^2-8x-4 = 0 # # :. 3(x^2-8/3x-4/3) = 0 # # :. (x^2-8/3x-4/3) = 0 # # :. (x-8/6)^2 -(8/6)^2-4/3 = 0 # # :. (x-8/6)^2 -64/36-4/3 = 0 # # :. (x-8/6)^2 = 64/36+48/36 # # :. (x-8/6)^2 = 112/36 # # :. (x-8/6) = +-sqrt(112/36) # # :. x=8/6 +-sqrt(112/36) # again leading to #x=-0.43, 3.10 # (2dp) The coordinates of these points are then given by using the original formula for #y# to give: # x=-0.43 => y=-16.9 # (1dp) # x=3.10 => y=5.0 # (1dp) These values of #x# correspond to the critical points (or turning points). we now use the second derivative to determine the nature (max, min or point of inflection) of these points. from earlier, # y'=-3x^2+8x+4 # # :. y''=-6x+8 # # x=-0.43 => y''=10.5 > 0 # (1dp) # x=3.10 => y''=-10.6<0 # (1dp) So we can now determine the nature of the turning points which are: # (-0.43,-16.9) # minimum # (3.10, 5.0) # maximum We now have enough to sketch the curve: graph{-(x-2)(x+2)(x-4) [-10, 10, -20, 20]}

Aurora Alvarado · Answer

undefined To sketch the graph of $ y = -(x-2)(x+2)(x-4) $ using the first and second derivatives, follow these steps:

1. **Find the First Derivative**:
   Calculate the first derivative $ y' $ to identify critical points (where $ y' = 0 $) and determine the intervals of increase and decrease.
   
   $$ y' = -1(x+2)(x-4) - (x-2)(x-4) - (x-2)(x+2) $$
   Simplify to:
   $$ y' = -(x^2 - 2x - 8) - (x^2 - 6x + 8) - (x^2 - 4) $$
   $$ y' = -x^2 + 2x + 8 - x^2 + 6x - 8 - x^2 + 4 $$
   $$ y' = -3x^2 + 8x - 4 $$

2. **Find Critical Points**:
   Solve for $ x $ when $ y' = 0 $.
   
   $$ -3x^2 + 8x - 4 = 0 $$
   Factor or use the quadratic formula to find the roots.

3. **Determine Intervals of Increase and Decrease**:
   Use test points from each interval to determine if the function is increasing or decreasing.

4. **Find the Second Derivative**:
   Calculate the second derivative $ y'' $ to identify points of inflection and concavity.
   
   $$ y'' = \frac{d}{dx}(-3x^2 + 8x - 4) $$
   $$ y'' = -6x + 8 $$

5. **Find Points of Inflection**:
   Solve for $ x $ when $ y'' = 0 $ to find potential points of inflection.

6. **Determine Concavity**:
   Use test points from each interval to determine if the function is concave up or concave down.

7. **Sketch the Graph**:
   - Plot the critical points, points of inflection, and any additional points determined from your tests.
   - Use the information from the intervals of increase/decrease and concavity to sketch the curve.
   - Ensure the curve passes through any intercepts or asymptotes.

Remember, this process provides a general sketch of the function. For more accurate details, plotting points or using graphing software can be beneficial.