Looking for someone who could possible tutor me in math on skype or something...please. I am in Alg. 2 and if I don't pass this year (I'm a senior in HS) I can't graduate...
Alot of us aren't very good at math myself included, so don't feel alone K? ;) Anyway I wished I could help but my brain just overheated on trying to understand it.
(−2, 18), (0, 4), (4, 24) okay we got this three points, a parabola in a function works the following always:
-------(Y = ax^2 + bx + c)-------- There has to be a way for the value X and Y in the point (0,4), 0=x and 4=Y
With that in mind we can have the following substitution on the parabola
We would have the following result -------(4 = a(0)^2 + b(0) + c)------by answering this the result would be : ~~(4=C)~~With this now we need the values of A and Bnow lets get the next point: (-2,18)by the substitution like last time we would get:------(18 = a(-2)^2 + b(-2) + c)------this would mean that our result would be the following:-----(18 = 4a -2b + c)----but lets remember we got C last time, so now it would become------(18=4a-2b+4)-----We send the value of 4 to next side, since its +4 it would become -418-4=4a-2b meaning we would end in ---(14=4a-2b)----so far we got two* 4=C* 14= 4a-2bnow lets go for the last point(4,24) by substitution we would get24 = a(4)^2 + b(4) + c by simplificating again we would get24 = 16a + 4b + 4, since we already knew the value of Cso in the end we would end up with----(20 = 16a + 4b)-----now we got two equations we can solve by system14= 4a-2band20= 16a+4bby system we can find lots of ways to know "A" and "B" I prefer this method(4a-2b=14)(2) we do this to eliminate the B16a+4b=20------------24a=48 would mean to getA=48/24A=2so now we can get b by simple substitution4(2)-2B=14-2B=14-8B=6/-2B=-3so our final result for the equation of the parabola in standard form would be :(Y = 2x^2 -3x + 4)in wich A=2, B=-3 and C=4well thats how you solve that one hope it helped you a bit
(−2, 18), (0, 4), (4, 24)
Solve by using tables. Give each answer to at most two decimal places.
x2 − 5x + 6 = 0
etc...a lot of them are pictures of graphs etc...I am not very smart at math.
a parabola in a function works the following always:
-------(Y = ax^2 + bx + c)--------
There has to be a way for the value X and Y
in the point (0,4), 0=x and 4=Y
With that in mind we can have the following substitution on the parabola
We would have the following result
-------(4 = a(0)^2 + b(0) + c)------by answering this the result would be : ~~(4=C)~~With this now we need the values of A and Bnow lets get the next point: (-2,18)by the substitution like last time we would get:------(18 = a(-2)^2 + b(-2) + c)------this would mean that our result would be the following:-----(18 = 4a -2b + c)----but lets remember we got C last time, so now it would become------(18=4a-2b+4)-----We send the value of 4 to next side, since its +4 it would become -418-4=4a-2b meaning we would end in ---(14=4a-2b)----so far we got two* 4=C* 14= 4a-2bnow lets go for the last point(4,24) by substitution we would get24 = a(4)^2 + b(4) + c by simplificating again we would get24 = 16a + 4b + 4, since we already knew the value of Cso in the end we would end up with----(20 = 16a + 4b)-----now we got two equations we can solve by system14= 4a-2band20= 16a+4bby system we can find lots of ways to know "A" and "B" I prefer this method(4a-2b=14)(2) we do this to eliminate the B16a+4b=20------------24a=48 would mean to getA=48/24A=2so now we can get b by simple substitution4(2)-2B=14-2B=14-8B=6/-2B=-3so our final result for the equation of the parabola in standard form would be :(Y = 2x^2 -3x + 4)in wich A=2, B=-3 and C=4well thats how you solve that one hope it helped you a bit