Use the graph of y=2/x^2 - x below to solve the equation 4x^3-10x^2+2=0.

Kulla_9289 said:
0 = (a + 1)x + b + 2 - 3x^2
So, 0 = -3x^2 + (a + 1)x + b + 2
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Yes, though you may find it helpful to write it as -3x^2 + (a + 1)x + (b + 2) = 0.

Now, you want this to be equivalent to 3x^2 – 7 = 0. Equate coefficients.
Kulla_9289 said:
Where is Cx? You didn’t include that in your calculations
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It turned out to be 0, which is why it was omitted in the advice someone gave you.

I take back my opinion that the new question belonged in this thread. I thought you were finished with the first, and understood it. Trying to talk about two things at once is unpleasant.
 
Dr.Peterson said:
Yes, though you may find it helpful to write it as -3x^2 + (a + 1)x + (b + 2) = 0.

Now, you want this to be equivalent to 3x^2 – 7 = 0. Equate coefficients.

It turned out to be 0, which is why it was omitted in the advice someone gave you.

I take back my opinion that the new question belonged in this thread. I thought you were finished with the first, and understood it. Trying to talk about two things at once is unpleasant.
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Don’t I need to change signs since I need it in the form of Ax^2 + Bx + c?
 
Kulla_9289 said:
Don’t I need to change signs since I need it in the form of Ax^2 + Bx + c?
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Yes. That shows you are thinking. (Though the reason is not what you say, but that the first coefficients need to be equal.)

Now do it!
 
Before I equate coefficients, the common term must be common (i.e. same signs). So, 3x^2 - (a + 1)x - (b + 2) = 0.
Equating coefficients: 3x^2 - (a + 1)x - (b + 2) = 3x^2 – 7
-(a + 1) = 0
-a - 1 = 0
-a = 0 + 1
-a = 1
a = -1
So, y = -1x + b
-(b + 2) = -7
-b - 2 = -7
-b = -7 + 2
-b = -5
b = 5
So, y = -x + 5
 
Kulla_9289 said:
@Dr.Peterson how would I solve 3x^2 – 7 = 0 if I am given a graph of y = 3x^2 – x – 2?
So, 3x^2 – x – 2 = ax + b.
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Kulla_9289 said:
Before I equate coefficients, the common term must be common (i.e. same signs). So, 3x^2 - (a + 1)x - (b + 2) = 0.
Equating coefficients: 3x^2 - (a + 1)x - (b + 2) = 3x^2 – 7
a = -1
b = 5
So, y = -x + 5
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So your answer is to intersect y = 3x^2 – x – 2 with y = -x + 5, which should give the solutions of 3x^2 – 7 = 0. Here are the three graphs:

1660411147194.png​

You can see that the intersections of the red and green are at the zeros of the red; so you're right.
 
@Dr.Peterson how would I solve 2^x = 2x + 2 if I am give a graph of y = 2^x?
So, ax + b = 2^x. But I need to manipulate it to the standard form of an exponential function (ab^x = 0).
So, (ax + b)2^x = 0?
 
Kulla_9289 said:
@Dr.Peterson how would I solve 2^x = 2x + 2 if I am give a graph of y = 2^x?
So, ax + b = 2^x. But I need to manipulate it to the standard form of an exponential function (ab^x = 0).
So, (ax + b)2^x = 0?
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Just take a moment to think.

The equation you want to solve is 2^x = 2x + 2. What intersection does that represent, just as it is?

You don't need to (and in fact can't) put it in the form you suggest; why would you do that? It's already in the exact form you need.
 
Kulla_9289 said:
@Dr.Peterson How would I solve 2^(x+2) = 40 if I am given a graph of y = 2^x?
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Please make an attempt and show us your work, as we ask you to do. (There are a couple different ways you could go.)

I have been helping you a lot; you should have learned enough to think for yourself at this point. Admittedly this one is a little different, but if you keep relying on others to do all the thinking, you will not learn to be independent.

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@Dr.Peterson So, ax + b = 2^x. I need to convert it into the form of a^x = 0. ax + b - 2^x = 0. I am not sure here since 2^x is already in its standard form.
 
Kulla_9289 said:
@Dr.Peterson So, ax + b = 2^x. I need to convert it into the form of a^x = 0. ax + b - 2^x = 0. I am not sure here since 2^x is already in its standard form.
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As I said, this one is a little different, so don't just put it into the same form you used for other problems when you find that it doesn't work. Try thinking differently. (Can you show us any examples you've been given, so we can see if other ideas besides those we've talked about might have been explained?)

One thing you might do is to rewrite the equation you want to solve, to change the feature of it that is most unlike what you've solved before, namely that the exponent is different than in the graph you are given.

Another approach would be to think about things you've learned about graphs. Have you learned about transformations of graphs? (Because we don't know any details about the course you are taking, or what has been taught recently, we can't tell what to suggest, so more such information might be helpful.)
 
No examples have been provided by the instructor, and I haven’t learnt transformation of graphs yet. Could you show an example on changing the feature of an equation I want to solve?
 
Kulla_9289 said:
No examples have been provided by the instructor, and I haven’t learnt transformation of graphs yet. Could you show an example on changing the feature of an equation I want to solve?
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Hint:
\(\displaystyle 2^{x+2}=2^x\times2^2\)
 
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