Question

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

Answer 1

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

Answer 2

4a = b

Answer 3

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

Answer 4

a = -3/4?

Answer 5

b = 3?

Answer 6

How did you get b = 2?

Answer 7

so a has different value too

Answer 8

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

Answer 9

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?

Answer 10

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

Answer 11

myininaya likes to go overboard with her methods

Answer 12

ok hero but there are two possible outcomes

Answer 13

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

Answer 14

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

Answer 15

Here I will do it the way I told you to

Answer 16

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

Answer 17

a = -3/4
b = -3

Answer 18

^did you make 2 equations and then substitutions?

Answer 19

something like that

Answer 20

-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? :)

Answer 21

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

Answer 22

I replace b with 4a because b = 4a

Answer 23

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

Answer 24

Which are all of the known values

Answer 25

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

Answer 26

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

Answer 27

\[-1=a(-2)^2+b(-2)-4\]
\[-1=4a-2b-4 => 3=4a-2b\]
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

Answer 28

:D

Answer 29

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

Answer 30

if a was squared, then there would be two possibilities

Answer 31

I didn't think about the y-intercept

Answer 32

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

Answer 33

yup

Answer 34

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

Answer 35

so my bad hero
you win this one

Answer 36

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

Answer 37

hey lady do you understand what i did up there?

Answer 38

A bit lol but thanks!

Answer 39

mya likes to confuse people then ask if they understand

Answer 40

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

Answer 41

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 :)

Answer 42

You're still here?