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lolab
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Erika's Equations | Math Solutions & Help
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SUMMARY
The discussion revolves around solving a system of equations derived from a quadratic function, specifically the equation \(y = f(x) = ax^2 + bx + c\). Dan outlines a method to solve for the coefficients \(a\), \(b\), and \(c\) using given points on the parabola, namely (1, 5) and (2, 10). A critical point raised is the inconsistency in having two different values for \(f(2)\), leading to the conclusion that no values of \(a\), \(b\), and \(c\) can satisfy both equations simultaneously, as shown by the contradiction when subtracting the equations.
PREREQUISITES- Understanding of quadratic functions and their standard form
- Basic algebraic manipulation skills
- Knowledge of solving systems of equations
- Familiarity with the concept of function values and graph interpretation
- Study methods for solving systems of linear equations
- Learn about the properties of quadratic functions and their graphs
- Explore the implications of inconsistent systems in algebra
- Practice deriving equations from given points on a parabola
Students studying algebra, educators teaching quadratic equations, and anyone interested in mathematical problem-solving techniques.
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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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