MHB Erika's Equations | Math Solutions & Help

[lolab]

View attachment 11839

 

[topsquark]

View attachment 11839

The link is bad.

-Dan

 

[lolab]

 

[topsquark]

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

 

[HallsofIvy]

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?