Hi everyone, I need some help with my Algebra homework. Any help is appreciated :) Find the equation in standard form of the parabola passing through the points. (3,-6), (1, -2), (6,3) For the second question it is the same thing. The points are (-1, -1), (1,3), and (-3,1) I am having a really hard time understanding how to do this. Thank you
Thank you, can you explain your steps to me so I can learn to do this on my own?
Is there anyone else able/available to help me through these questions?
with 3 random points, one method would be to define 3 equations in 3 unknowns
a parabola takes the form: ax^2 + bx + c = y we know the x and y parts, it is the abc parts that are unknown to us
Ok
(3,-6), (1, -2), (6,3) when x=3 and y=-6 a(3)^2 + b(3) + c = -6 9a + 3b + c = -6 --------------------------- when x=1 and y=-2 a(1)^2 + b(1) + c = -2 a + b + c = -2 --------------------------- when x=6 and y=3 a(6)^2 + b(6) + c = 3 36a + 6b + c = 3 --------------------------- this forms a system of 3 equations; in 3 unknowns 9a + 3b + c = -6 a + b + c = -2 36a +6b + c = 3
there are different methods that can be used to find the abc values .. substitution, elimination, and matrix methods tend to be used the most
Ok, so if we use elimination what would happen next?
you would use the elimination method then ... are you familiar with it and just unsure how to use it for a 3 equation setup?
I am familiar with it, just a little bit confused.
we wold use one of the equations to eliminate a variable out of the other 2 .. instead of just out of "the other one". Since the middle equation is fairly naked, it looks like a good one to use: scale it by some value k, to eliminate an unknown ka + kb +kc = -2k 9a + 3b + c = -6 scale it by some value n, to eliminate the same unknown in the other equation na + nb + nc = -2n 36a +6b + c = 3 i spose the question now is, which variable do you want to eliminate first?
I guess c
then id say let k = -1, and let n = -1 and add; what equations do we end up with?
I am a little bit confused with the way you set up the equations for elimination. I'm sorry
-a - b - c = 2 9a +3b + c = -6 ------------------- 8a +2b = -4 -a - b - c = 2 36a +6b + c = 3 ------------------ 35a +5b = 5
i simply picked one of the equations to modify in order to eliminate a variable from the other 2
that is all we did with a system of 2 equations in 2 unkowns; pick one and modify it to eliminate a variable in the other equation
Ok, it is starting to make more sense
the elimination we just did creates 2 equations in 2 unknowns .. which can be worked again 8a +2b = -4 35a +5b = 5 this should be something you are more used to dealing with
Yes, so that answer is 43a+7b = 1
no, we want to run another elimination on the system we created. im going to reduce the coefficients to make the numbers smaller 8a +2b = -4 reduce by dividing off 2 35a +5b = 5 reduce by dividing off 5 4a +b = -2 7a +b = 1 b looks like an easy target to eliminate; multiply the first equation by -1 -4a -b = 2 7a +b = 1 ------------- 3a = 3 solving for a is simple now; a = 1
Oh, Ok. Got it.
now that we know a, take an ab setup to solve for b 7a +b = 1 7 +b = 1 b = -6 a = 1, b = -6 now that we know ab, take an abc setup to solve for c a + b + c = -2 1 - 6 + c = -2 -5 + c = -2 c = 3 we now know all the abc parts to fill in the quadratic: ax^2 + bx + c = y x^2 -6x +3 = y
Yea! thank you so much for helping me! It makes a lot more sense now
youre welcome; that is what we are spose to be here for instead of like that first poster did ... just giving out an answer has no learning value :/
No, I need to be able to work things out myself. Thank you for helping :)
i have class starting soon so i have to go, good luck with this
Have a great day
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