Question

Find real numbers a, b, and c so that the graph of the quadratic function y=ax^2+bx+c contains the points (-2, 13), (0,1), and (-2,3)

Answer 1

Might have to solve a system of linear equations here. You cool with that?

Answer 2

yes! i've solved some of this already, but it seems inconsistent to me once I build the second set of equations. I'm ready! :)

Answer 3

Ok. Set up the equations, you have three of them, but one will dissolve quite nicely since it is the y-intercept.

y=ax^2+bx+c
(-2, 13): 13=a(-2)^2+b(-2)+c
(0,1): 1=a(0)^2+b(0)+c
(-2,3): 3=a(-2)^2+b(-2)+c

Hmm, yeah, I see what you're saying about being inconsistent. Are you sure you have the points right? There shouldn't be two y-values for a single x input..

Answer 4

Maybe this is supposed to be x=ay^2+by+c? Have you done general quadratics and conic sections yet? This could be a 'sideways' parabola.

Answer 5

i attached the actual page for you to view

Answer 6

I can't open Powerpoint files (especially ones with embedded macros - my antivirus software will throw a fit over that).

Answer 7

ok, in that case, yes, i'm sure that's the exact problem. word for word.

Answer 8

i got it down to the original 3 equations. 4a-2b+c=13....c=1....4a-2b+c=3. That's where i multiplied them through, eliminating c, and ended up with 4a-2b=12 and 4a-2b=2. Is my process up to that point correct? Because that's when I end up with 0=10 and just can't continue. Is there an approach on the calculator I could utilize? with the matrix option?

Answer 9

I don't think matrices are necessary. It might be a degenerate quadratic that reduces to a single straight line, but the points don't seem to add up. I'll check it again.

Answer 10

parallel perhaps? that would support the inconsistent status atleast

Answer 11

so it's not just me.....phew!

Answer 12

Yeah, the system reduces to
2a-b=6
2a-b=1
which is inconsistent for a and b, but we can at least trust c=1.

Maybe the equation is simply y=1 and is a horizontal line.

Answer 13

It doesn't work for the other two points though, which means it can't be a function of x.
It could be a function of y, though. Try the sideways parabola.

Answer 14

i'm not sure what you mean. we just learned this on wednesday, new material, but this was the problem our teacher challenged us to over the weekend. hmmm

Answer 15

we begin matrices on monday, does that guide you at all here?

Answer 16

Nope. You can use matrices if you want, but it will yield exactly the same results.
The system reduces to a 2x2, which is easy to do with matrices (especially with the Cramer's Rule short cut, but is completely unnecessary).

Try graphing the thing. Plot the points and try drawing a parabola through them.

Answer 17

ok

Answer 18

eh, it's more like half of a x^2+1graph, what does that mean? definitely not going to be sideways.

Answer 19

Ah, then you can ignore one of the points and only use the first quadrant part.

Answer 20

wouldn't it be quadrant II?

Answer 21

Yeah, had to use my imagination a little more.
Here's the sideways parabola idea: