# How do you use the first and second derivatives to sketch #y = (x+2)(25 -x^2)#?

I got:

and Socratic graphs show:

graph{(x+2)(25 - x^2) [-8, 6, -107.6, 107.8]}

*DISCLAIMER: LONG ANSWER!
*

*To graph this, you need:*

- All
#x# intercepts - Coordinates for all extrema
- What these extrema are (maximum or minimum)
GETTING THE X INTERCEPTS

You fortunately have

#y# mostly factored.#color(green)(y) = -(x+2)(x^2 - 25)# #= color(green)(-(x+2)(x+5)(x-5))# ---*difference of two squares*Right now, if you set this equal to

#0# , you'll find that the roots are:#x = -5# , or#color(blue)("("-5,0")")# #x = -2# , or#color(blue)("("-2,0")")# #x = 5# , or#color(blue)("("5,0")")#

The three roots tell you that there are three

#x# intercepts and the polynomial is a cubic.GETTING THE COORDINATES FOR ALL EXTREMA

Now multiply it all out to take the derivatives more easily using the power rule.

#-(x+2)(x^2 - 25)# #= -(x^3 - 25x + 2x^2 - 50)# #= -x^3 - 2x^2 + 25x + 50# Now the first derivative, which gives you where the

*extrema*are, is:#color(green)((dy)/(dx))# #= d/(dx)[-3x^2 - 4x + 25]# #= color(green)(-3x^2 - 4x + 25)# When you set this equal to

#0# and solve, you are looking for where the graph*changes direction*, which is where its slope is#0# .Because you're solving a quadratic, you should get two roots that each tell you where the extrema are. But you

*don't*know whether they are minima or maxima yet. The*second*derivative tells you that.Unfortunately this doesn't factor into integer solutions, so we'll have to

*complete the square*.- Get the
#x^2# term coefficient to be#1# . - Take the resultant
#x# term, halve the coefficient, and square the coefficient to get the#x^0# term. - Make sure you've added the resultant
#x^0# term to both sides.#-3x^2 - 4x = -25# #3x^2 + 4x = 25# #3(x^2 + 4/3x) = 3(25/3)# #3(x^2 + 4/3x + 4/9) = 3(25/3 + 4/9)# #3(x + 2/3)^2 = 79/3# #=> color(green)((dy)/(dx) = 3(x + 2/3)^2 - 79/3 = 0)# Now, solve this for

#x# to get:#(x+2/3)^2 = 79/9# #x + 2/3 = pmsqrt(79/9)# #color(green)(x_1 = sqrt(79/9) - 2/3 ~~ 2.30)# #color(green)(x_2 = -sqrt(79/9) - 2/3 ~~ -3.63)# If we plug these in back to the original equation, we can get the coordinates for the extrema.

#color(green)(y_1) = -x_1^3 - 2x_1^2 + 25x_1 + 50 ~~ color(green)(84.75)# #color(green)(y_2) = -x_2^3 - 2x_2^2 + 25x_2 + 50 ~~ color(green)(-19.27)# So the two extrema in the graph are:

#color(blue)( "("2.30,84.75")")# #color(blue)( "("-3.63,-19.27")" )#

Now to figure out which one is the maximum and which one is the minimum.

DETERMINING WHETHER THEY'RE MAXIMA OR MINIMA

With the second derivative, we have three options:

- If
#|[(d^2y)/(dx^2)]|_(x = a) > 0# , we have a minimum. - If
#|[(d^2y)/(dx^2)]|_(x = a) < 0# , we have a maximum. - If
#|[(d^2y)/(dx^2)]|_(x = a) = 0# , we have an inflection point, where the concavity changes from concave up to concave down or vice versa.We have to select the

#x_1# or#x_2# values we found just now since those correspond to the only two extrema.#color(blue)(|[(d^2y)/(dx^2)]|_(x ~~ 2.30))# #= d/(dx)[(dy)/(dx)]# #= d/(dx)[-3x^2 - 4x + 25]# #= -6x - 4# #=> -6(2.30) - 4 ~~ color(blue)(-17.78 < 0)# #=># #\mathbf(color(blue)("maximum"))# (inside of curve faces down)#color(blue)(|[(d^2y)/(dx^2)]|_(x ~~ -3.63))# #=> -6(-3.63) - 4 = color(blue)(17.78 > 0)# #=># #\mathbf(color(blue)("minimum"))# (inside of curve faces up)There, now we have enough information to sketch this!

GRAPHING THE RESULT

All our coordinates that we just solved for are...

x-intercepts

#(-5,0)# #(-2,0)# #(5,0)#

extrema

#(-3.63, -19.27)# ---*minimum*#(2.30, 84.75)# ---*maximum*

So now we graph this as:

and Socratic graphs show:

graph{(x+2)(25 - x^2) [-8, 6, -107.6, 107.8]}

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*Answer 2Sign up to view the whole answerSign up with email*To sketch the graph of ( y = (x+2)(25 -x^2) ) using the first and second derivatives:

- Find the first derivative ( y' ) using the product rule.
- Find the critical points by setting ( y' = 0 ) and solving for ( x ).
- Determine the intervals where the first derivative is positive or negative to identify increasing and decreasing portions of the graph.
- Find the second derivative ( y'' ) to identify concavity.
- Determine the intervals where the second derivative is positive or negative to identify concave up and concave down portions of the graph.
- Plot the critical points, inflection points, and other key points.
- Sketch the graph accordingly, considering the behavior of the function based on the signs of the derivatives.

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*Answer from HIX Tutor**When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

*When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.*

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