Question

Determine the quadratic equation for the parabola:

minimum value: -5; x-intercepts: -2 and 3

Answer 1

y = a(x+2)(x-3)

Answer 2

get dy/dx = 0
set the minimum value =-5 and find a

Answer 3

i might be off on that lol

Answer 4

f(x)=a(x-h)^2+k
vertex is (n,-5)
x-intercepts (-2,0) and (3,0)
f(-2)=0
f(3)=0
f(h)=-5=k
f(-2)=a(-2-h)^2+k=0
f(3)=a(3-h)^2+k=0
a(4+2*2h+h^2)+k=0
a(h^2+4h+4)+k=0 => a(h^2+4h+4)=-k
f(3)=a(3-h)^2+k=0
a(9-2*3h+h^2)+k=0
a(9-6h+h^2)+k=0 => a(9-6h+h^2)=-k
so
a(h^2+4h+4)=a(9-6h+h^2)
h^2+4h+4=h^2-6h+9
4h+6h=9-4
10h=5
h=2
h=2
now we need to know a?
a(3-h)^2+k=0
a(3-2)^2-5=0
a-5=0
a=5
so f(x)=5(x-2)^2+5

Answer 5

f(x)=5(x-2)^2-5*

Answer 6

wait my (-2,0) is not working
:(

Answer 7

get the generic family by (x+2)(x-3); the point in the middle of your zeros is the vertex :) so use that point to calibrate this to -5

Answer 8

-2+3 = 1; /2 = 1/2

the point (1/2,y) should equal be (1/2,-5) when youre done

Answer 9

omg i see my mistake

Answer 10

im such a tard
10h=5
h is not 2
h is 1/2

Answer 11

(1/2)^2 -(1/2) -6 = at x=.5; we get -6.25; -5/-6.25 = the scalar perhaps?

Answer 12

I get .8 if its right; now multiply everything in the equation by .8 to see if it fits :)

Answer 13

a(3-1/2)^2-5=0
a(3-1/2)^2=5
a(5/2)^2=5
a(25/4)=5
a=4/25*5
a=20/25
a=4/5
so f(x)=4/5*(x-1/2)^2-5

Answer 14

.8 x^2 -.8 x -4.8 is what I get :)

Answer 15

i win :) my works does yours

Answer 16

your x^2 coefficent is right...

Answer 17

looks good amistre :)

Answer 18

mine works lol; I just tested it :)

Answer 19

i think i approached this from 3 blocks away when i could had approached it if i just stayed in my own house

Answer 20

you like my metaphor

Answer 21

?

Answer 22

it ok; needs a rocket tho lol

Answer 23

lol

Answer 24

later