Question

the parabola has a y-intercept of 5 anf contains the point (2,-3) and (-1,12). what are the coordinates of the vertex?

Answer 1

(0,5) (2,-3) (-1,12) right?

Answer 2

yes

Answer 3

0a + 0b + 5 = 5

4a +2b + 5 = -3

a -b +5 = 12 ; a = 12-5 +b
----------------------------
a = 7 +b
----------------

4(7+b) +2b + 5 = -3

28 +4b +2b = -8

6b = -36

b = -6
c = 5
-----------------------

a = 12-5 +b; a = 12-5-6

a = 1
--------------------------

y =x^2 -6x +5 is your equation for hese points

Answer 4

the vertex is at -b/2a

6/2 = 3

the vertex is at (3,y)

y = (3)^2-6(3) +5

y = 9 -18 +5

y = 14 -18

y = -4

vartex is at (3,-4) if I did it right ;)

Answer 5

you agree or disagree :)

Answer 6

thanx ..that is the rite answer
i
agree

Answer 7

yay!! if you ever wanna know "how" its the right answer... I can do that too :)

Answer 8

ya..i dont get the first few steps

Answer 9

we know that these three points are not in a straight line, so that only leaves a parabola as an answer that we can actually get.

Answer 10

a parabola usually is of the form:

ax^2 +bx + c = y ; and you normally try to figure out the "x" and "y" values...but they already give you those right?

Answer 11

yup

Answer 12

so we get the first 3 equations:

a(0)^2 +b(0) +c = 5 ; (0,5).. c = 5 naturally

Answer 13

the "variables" we are trying to get them to have on common are the a, b, and c parts so this is almost a backwards solution lol

Answer 14

a(2)^2 + b(2) + c = -3

4a +2b +5 = -3

4a +2b = -8 right?

Answer 15

yup..

Answer 16

a(-1)^2 +b(-1) + c = 12 ; (-1,12)

a -b +5 = 12

a - b = 12 - 5

a = 7 +b right?

Answer 17

oh..i get it!!

Answer 18

4a +2b = -8 ; so plub if our "value" for "a"

4(7+b) +2b = -8

28 +4b +2b = -8

6b = -36

b = -6 :)

Answer 19

a = 7+b so...

a = 7 -6 = 1

our quad equation becomes:

x^2 -6x +5 = y :)

Answer 20

thanx a lot friend..

Answer 21

youre welcome :)

Answer 22

would u like to look through the other question i posted too..

Answer 23

if I can find it lol

Answer 24

okay