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lolab
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MHB Erika's Equations | Math Solutions & Help
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The discussion centers around solving a set of equations derived from a quadratic function, specifically addressing the impossibility of having two different outputs for the same input. The equations are based on points (1, 5) and (2, 10) on the graph of the function, leading to a contradiction when a third point (2, 19) is introduced. This contradiction arises because both equations for the value at x=2 yield the same left-hand side, making it impossible for them to equal different right-hand sides. The suggested approach involves solving the equations step-by-step, but the underlying issue of conflicting outputs remains a critical point of discussion. The conclusion emphasizes the impossibility of finding values for a, b, and c that satisfy the conflicting equations.
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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?
Hi,
I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance!
Question 1:
Around 4:22, the video says the following.
So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
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