Modelling with quadratics

[worn horizon]

How do i get a -

[candid cairn]

Yo so what have you tried

Do you have any ideas to start

[worn horizon]

i tried doing

i did

a(H-40)^2+12

1600a=-12

a=-0.0075

-0.0075(H-40)^2+12

[candid cairn]

Also by your question, are you asking how to get part a? Just checking you aren’t asking about negatives

[worn horizon]

😔 but its wrong-

[candid cairn]

Yep so you do havw the right idea

[worn horizon]

yeah how to get part a

[candid cairn]

So the bit where you said a(H-40)^2 + 12

Tell me how you decided on this, where’s the 40 and 12 from

[worn horizon]

i just realized the y intercept is supposed to be 0 not 12

[worn horizon]

from (40,0) and (0,12) where the ball lands and it's maximum height

[candid cairn]

Ah yeah (0,12) isn’t quite right

[worn horizon]

wait how 😭

[candid cairn]

Because the max height is achieved not when x = 0

It’s at that point in the middle right?

[worn horizon]

yeah

[worn horizon]

oh-

[candid cairn]

So what’s the x value there in the middle

Where the max height happens

[worn horizon]

x=20

[candid cairn]

Ok awesome, do you still want my help or do you wanna try work from here solo?

(40,0) and (20,12) are points the quadratic goes through (and (0,0))

[worn horizon]

I think i might be able to work it out! Thank you so much mister monster!!

[candid cairn]

No worries Kruyzfelz!