Graphing Quadratic Functions: How to Find Critical Points and Inflection Points

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Homework Help Overview

The discussion revolves around graphing quadratic functions, specifically focusing on finding critical points and inflection points for the function y = x²(x² - 4)². Participants are exploring the necessary steps to analyze the function for graph sketching.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the process of finding critical points and inflection points, with some attempting to differentiate the function and solve for roots. Questions arise regarding the application of the quadratic formula and the identification of even or odd functions.

Discussion Status

The conversation is ongoing, with participants providing hints and guidance on how to approach solving for inflection points and clarifying the nature of the function. There is acknowledgment of errors in algebraic manipulation, and some participants express a desire to work through the problem independently.

Contextual Notes

There is mention of a typographical error in the original function, which has led to confusion in the analysis. Participants are also reflecting on their understanding of function properties without complete reliance on algebraic expansion.

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Homework Statement


I've been looking at the problem for a while and I just can't figure it out. It is graph sketching.

[STRIKE]y=x2(x2-4)2[/STRIKE]
the correct equation is this:
y=x2(x-4)2

Homework Equations



Solve for critical points, and inflection points in order to get points needed to graph function.

The Attempt at a Solution



[STRIKE]y=x2(x-4)2[/STRIKE]
y=x2(x2-4)2
y=x4-8x3+16x2

y'=4x3-24x2+32x
0=4x(x-4)(x-2)
x=0,2,4

local min: 0,4
local max: 2

y"=12x2-48x+32

I don't know what to do to solve for the inflection points.
 
Last edited:
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MAC5494 said:
I don't know what to do to solve for the inflection points.
Well, inflection points can be arrived at by solving f "(x)=0 and finding values of x and substituting them again in the given equation.
That should give you all inflection points of that function.
 
vhbelvadi said:
Well, inflection points can be arrived at by solving f "(x)=0 and finding values of x and substituting them again in the given equation.
That should give you all inflection points of that function.

Yeah I'm aware of that, and I've tried, but I can't figure out how to solve for x.
 
MAC5494 said:
Yeah I'm aware of that, and I've tried, but I can't figure out how to solve for x.

Are you asking us how you might solve ##12x^2 - 48x + 32 = 0##?
 
FeDeX_LaTeX said:
Are you asking us how you might solve ##12x^2 - 48x + 32 = 0##?

Yes, or at least some pointers in the right direction. I'd like to try to do it myself if possible.
 
MAC5494 said:
Yes, or at least some pointers in the right direction. I'd like to try to do it myself if possible.

Have you seen the quadratic formula before?

In relation to sketching the original curve, it might save you some time to note whether or not the function is odd or even.
 
FeDeX_LaTeX said:
Have you seen the quadratic formula before?

In relation to sketching the original curve, it might save you some time to note whether or not the function is odd or even.

Yes! I didn't even think about that! Thank you for your help! It should be an even function just by looking at the formula. How do you figure out algebraically if it is even or odd?
 
MAC5494 said:
Yes! I didn't even think about that! Thank you for your help! It should be an even function just by looking at the formula. How do you figure out algebraically if it is even or odd?

In general, a function ##f(x)## is odd if ##f(-x) = -f(x)##, and even if ##f(-x) = f(x)##. (You can also decompose a function into an even part and an odd part, but that's irrelevant here.)

What happens when you replace ##x## with ##-x## in your function?
 
FeDeX_LaTeX said:
In general, a function ##f(x)## is odd if ##f(-x) = -f(x)##, and even if ##f(-x) = f(x)##. (You can also decompose a function into an even part and an odd part, but that's irrelevant here.)

What happens when you replace ##x## with ##-x## in your function?

f(x) = x4-8x3+16x2

f(-x) = (-x)4 -8(-x)3+16(x)2

= x4 + 8x3 + 16x2

So I was wrong. It's actually neither even or odd.
 
  • #10
From the OP:
y = x^{2}(x^{2}-4)^{2}
y = x^{4}-8x^{3}+16x^{2}

You need to work on your basic algebra skills: the first equation does not lead to the second.

y = x^{2}(x^{2}-4)^{2}
y = x^{2}(x^{4}-8x^{2}+16)
y = (x^{6}-8x^{4}+16x^{2})
 
  • #11
SteamKing said:
From the OP:
y = x^{2}(x^{2}-4)^{2}
y = x^{4}-8x^{3}+16x^{2}

You need to work on your basic algebra skills: the first equation does not lead to the second.

y = x^{2}(x^{2}-4)^{2}
y = x^{2}(x^{4}-8x^{2}+16)
y = (x^{6}-8x^{4}+16x^{2})

Oh god. Well there's my issue. I forgot to foil. Thanks for your help. I don't think I'll have an issue with this problem anymore. That's embarrassing.
 
  • #12
MAC5494 said:
f(x) = x4-8x3+16x2

f(-x) = (-x)4 -8(-x)3+16(x)2

= x4 + 8x3 + 16x2

So I was wrong. It's actually neither even or odd.

The function is even; x^2 is even and x^2 - 4 is even, so x^2 * (x^2-4)^2 is even. No work needed!
 
  • #13
Ray Vickson said:
The function is even; x^2 is even and x^2 - 4 is even, so x^2 * (x^2-4)^2 is even. No work needed!

I forgot to foil so I came out with the wrong function.
 
  • #14
MAC5494 said:
I forgot to foil so I came out with the wrong function.

Yes, but you can tell that the function is even without any expansion, since you have a product of two even functions.
 
  • #15
FeDeX_LaTeX said:
Yes, but you can tell that the function is even without any expansion, since you have a product of two even functions.

Yeah I know. I said above that it should be even. However when I actually did it, I used the wrong function, which was the function I for some reason didn't foil. I appreciate everyone's help.
 
  • #16
This is going to sound funny, but I typed in the function above wrong. I didn't sleep much last night as you can probably tell.

y=x2(x2-4)2

It is supposed to be y=x2(x-4)2

So I have the same problem. I'll try doing the quadratic formula to see what I get.
 

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