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first simplify the function by taking out -16
so
y = -16(t^2 + 2t - 24 )

can you factor the quadratic in the parentheses?

yes hang on one sec please

f(t)=−16(t+6)(t−4)

So thats it for part A?

wait what would be the meaning???

t is seconds?

- there will be 2 x-intercepts

t is seconds yes

wait im lost sorry

ok so Part A would be f(t)=−16(t+6)(t−4) where t is equaled to seconds?

f(t) = -16(t + 6)( t - 4) = 0
find values of t

- the x intercepts are values of t when f(t) the height = 0.

So the x-intercepts or t are equaled to -6 and 4?

and the x-intercept is (4,0) which represents the time in seconds that the sandbag was thrown?

to where it lands

for Part b the completing of the square is: (t+1)^2=25 right

yes t = 0 is the time when in is thrown and t = 4 = when it lands

don't forget the -16

ok so part a is complete!

Wait the -16 for what??

part a is complete yes
the function is -16(t^2 + 2t - 24)

the function is f(t)=−16(t+6)(t−4
For Part B: complete the square: (t+1)^2=25

-16(t^2 + 2t - 24)
= -16 [ (t + 1)^2 - 15]

now I have to find the vertex and decide whether it is a maximum or minimum

= -16 [ (t + 1)^2 - 25]

yes it will be a maximum because the coefficient of t^2 is negative

ok so (t+1)^2=25 is a maximum with the vertex or 400??

of^^^

excuse me i 'll be back in a minute or two...

ok

the axis of symmetry is the vertical line passing through the point (-1,400) which is x = -1

* t = -1

ok I understand thanks so much! Is that everything??

ok thank you! @welshfella I may need more help on other questions Ill tag you!!!!!!!!

ok yw