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lolab
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Erika's Equations | Math Solutions & Help
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Discussion Overview
The discussion revolves around solving a set of equations related to a quadratic function, specifically addressing the implications of given points on the graph of the function. Participants explore methods for solving the equations and raise concerns about the validity of the provided points.
Discussion Character
- Homework-related
- Mathematical reasoning
- Debate/contested
Main Points Raised
- One participant, Dan, questions the validity of a link and offers a method for solving a system of equations involving three unknowns.
- Another participant asserts that the problem is impossible due to conflicting values for the function at the same input, specifically pointing out that the equations derived from the points (2, 10) and (2, 19) lead to a contradiction.
- The participant further elaborates on the implications of having two different values for the same input in a function, prompting a discussion on the feasibility of finding values for a, b, and c that satisfy the equations.
Areas of Agreement / Disagreement
Participants express disagreement regarding the validity of the equations derived from the points on the graph, with one participant asserting the impossibility of the situation while another focuses on the method of solving the equations.
Contextual Notes
The discussion highlights potential limitations in the problem setup, particularly regarding the assumptions about the values of the function at specific points and the implications of having multiple outputs for a single input.
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The link is bad.lolab said:
-Dan
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lolab
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Do you understand how they got the first two equations? Can you get part a)?
As to the second part you have three equations in three unknowns. I'd solve the first equation for c, then plug that into the other two equations. Then solve one of them for b and plug that into the next. Then solve the last equation for a.
See what you can do with this and if you are still having problems post what you've got and we can take a look at it.
-Dan
As to the second part you have three equations in three unknowns. I'd solve the first equation for c, then plug that into the other two equations. Then solve one of them for b and plug that into the next. Then solve the last equation for a.
See what you can do with this and if you are still having problems post what you've got and we can take a look at it.
-Dan
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HallsofIvy
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This is immediately impossible. f(2) cannot have two different values!
We are given $y= f(x)= ax^2+ bx+ c$. The fact that (1, 5) is on its graph (a parabola) means that $y= 5= a(1)^2+ b(1)+ c= a+ b+ c$.
The fact that (2, 10) is on the graph means that $y= 10= a(2)^2+ b(2)+ c= 4a+ 2b+ c$.
If (2, 19) were also on the graph we would have $y= 19= a(2)^2+ b(2)+ c= 4a+ 2b+ c$.
So we have both $10= 4a+ 2b+ c$ and $19= 4a+ 2b+ c$.
What do you get if you subtract the first equation from the second? Is there ANY value of a, b, and c which will make that true?
We are given $y= f(x)= ax^2+ bx+ c$. The fact that (1, 5) is on its graph (a parabola) means that $y= 5= a(1)^2+ b(1)+ c= a+ b+ c$.
The fact that (2, 10) is on the graph means that $y= 10= a(2)^2+ b(2)+ c= 4a+ 2b+ c$.
If (2, 19) were also on the graph we would have $y= 19= a(2)^2+ b(2)+ c= 4a+ 2b+ c$.
So we have both $10= 4a+ 2b+ c$ and $19= 4a+ 2b+ c$.
What do you get if you subtract the first equation from the second? Is there ANY value of a, b, and c which will make that true?
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