how to determine a & b valuesin the equation: y=ax^2 +bx-4 if the vertex is located at (-2,-1) Please help!

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- myininaya

did you enter in -2 for x and -1 for y

- Hero

4a = b

- Hero

-1 = a(-2)^2 +(4a)(-2) - 4

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## More answers

- Hero

a = -3/4?

- Hero

b = 3?

- Hero

How did you get b = 2?

- myininaya

so a has different value too

- Hero

myininaya, how come you just can't do things the normal way?

- anonymous

so you cant just sub -2 into one equation and -1 into another, andthe other number equal zero? or does that not mke any sense?

- anonymous

I'm just a little confused by the math you did above..lol

- Hero

myininaya likes to go overboard with her methods

- myininaya

ok hero but there are two possible outcomes

- Hero

Not if you do it the way I did. I used direct substitution

- myininaya

We have a parabola concave up and also concave down at vertex

- myininaya

Here I will do it the way I told you to

- anonymous

Im familar with substitution , just not the way shown above im only in gr 10 so mabye thats why i dont get yours?

- Hero

a = -3/4
b = -3

- anonymous

^did you make 2 equations and then substitutions?

- Hero

something like that

- anonymous

-1 = a(-2)^2 +(4a)(-2) - 4 :this is what you did, why did you sub 4a into the equation? is there a more strait forward way to do this? :)

- Hero

myininaya, if you use the direct substitution method, you wouldn't have to worry about getting hung up like that. The parabola opens upward or downward depending on the value of a. Since I calculated a to be negative, I'm pretty sure it opens downward and there is only one possibility

- Hero

I replace b with 4a because b = 4a

- Hero

I also replaced x = -2 and y = -1

- Hero

Which are all of the known values

- Hero

according to x = -b/2a, there's only one possible a

- anonymous

alright im still a bit confused but thanks for your help!! :)

- myininaya

\[-1=a(-2)^2+b(-2)-4\]
\[-1=4a-2b-4 => 3=4a-2b\]
We also know the vertex which is (-2,-1)
\[y=a(x-h)^2+k\]
we also know h=-2 and k=-1
so we have
\[y=a(x+2)^2-1=a(x^2+4x+4)-1=ax^2+4ax+4a-1=ax^2+4ax+(4a-1)\]
\[=>4a=b , -4=4a-1\]
\[=>3=b-2b=> 3=-b=> b=-3\]
if b=-3 then a=-3/4
hmmm how do we get the other possibility
ok so maybe there is only one lol

- Hero

:D

- Hero

There's only one possibility because there's only one a value that would work in order to get both sides equal for x = -b/2a

- Hero

if a was squared, then there would be two possibilities

- myininaya

I didn't think about the y-intercept

- myininaya

The parabola has vertex (-2,-1) and -4 is lower so this graph will be concave down

- myininaya

|dw:1327278866372:dw|

- Hero

yup

- myininaya

we can't not have it concave up is has to come to that -4

- myininaya

so my bad hero
you win this one

- Hero

I don't win many so I will savor this :P

- myininaya

hey lady do you understand what i did up there?

- anonymous

A bit lol but thanks!

- Hero

mya likes to confuse people then ask if they understand

- myininaya

you also have to use the vertex form like i did to find another equation or two

- anonymous

im still pretty confused but i can move on to a different qestion on my assignment lol thanks for the help on this one tho :)

- Hero

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