Plz help
use The information provided to answer the following questions

Part A

By evaluating the derivative with respect to t of the height function h(t),
which is h’(t)=-2t+4,

we can find the turning point of the quadratic graph by solving h’(t)=0.

-2t+4=0
t=2

Therefore, we can know that when t=2, Sally’s potato attains its maximum height, which is
-(2^2)+4(2)+5
=9
Given that Sam’s potato has been 10 unit high, Sam’s potato went higher than Sally’s did.



Part B

Note that the for all t where h(t)>0, the potato is still in the air.
So all we need to do is to solve h(t)>0.

-t^2 + 4t + 5 > 0
t^2 - 4t - 5 < 0
(t + 1)(t - 5) < 0
Therefore we get -1 < t < 5.

But t represents time so it can’t be negative.
So the overall solution is 0 < t < 5.
Therefore the total flying time of Sally’s potato is 5 units.

Because the flying time of Sam’s potato is 3.5 units,
Sally’s potato stayed in the air longer.

Is there a picture of a graph or whatever that goes along with this?

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph

there is a hike in both the graphs so the graphs are increasing on their domain.x-intercept of log x and log 6x are differentthere is a vertical asymptote of both the graphs so there is no y-interceptas both the graphs are increasing so there is no horizontal asymptote