Stuck on this. 60 points!

When this graph is plotted and you imagine the x axis to be time (as stated in the question), each of the roots (x - intercept) must be when the swimmer goes under and when they come back up.

This means that the swimmer dived under the water at 5 seconds and came back up at 9, making the first 2 statements correct.

Now the fourth statement is ruled out.

The fifth statement is not plausible as the graph would have to have more than 2 roots for the swimmer to enter the water twice.

That leaves the third statement. If you imagine the depth of the swimmer to be the y axis of our imaginary graph, and we know that the y axis of the turning point is -9, that means that the swimmer's deepest dive was 9 feet under the water, ruling out the third statement too.

Hope this helps :D

semsee45 semsee45 · Answer 2 · 2022-06-28T13:35:08+02:00

Answer:

[tex]h(t)= (t-5)(t-11)[/tex]

The swimmer comes back up 11 seconds after the timer is started.

The swimmer dives into the water 5 seconds after the timer is started.

Step-by-step explanation:

Given function:

[tex]h(t)=t^2-16t+55[/tex]

To factor a quadratic in the form [tex]ax^2+bx+c[/tex],

find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex]:

[tex]\implies ac=55[/tex]

[tex]\implies b=-16[/tex]

Two numbers that multiply to 55 and sum to -16 are: -11 and -5

Rewrite b as the sum of these two numbers:

[tex]\implies t^2-11t-5t+55[/tex]

Factorize the first two terms and the last two terms separately:

[tex]\implies t(t-11)-5(t-11)[/tex]

Factor out the common term (t - 11):

[tex]\implies (t-5)(t-11)[/tex]

Therefore, the given formula in factored form is:

[tex]h(t)= (t-5)(t-11)[/tex]

The swimmer's depth is modeled as h(t). Therefore, when h(t) = 0 the swimmer will be at the surface of the water.

[tex]\implies h(t)=0[/tex]

[tex]\implies (t-5)(t-11)=0[/tex]

[tex]\implies t-5=0\implies t=5[/tex]

[tex]\implies t-11=0 \implies t=11[/tex]

Therefore, the swimmer will be at the surface of the water at 5 s and 11 s.

The swimmer's maximum depth is the vertex of the function. The x-value of the vertex is the midpoint of the zeros. Therefore, the x-value of the vertex is t = 8.

Substitute t = 8 into the function to find the maximum depth:

[tex]\implies h(8)=8^2-16(8)+55=-9[/tex]

So the swimmer's maximum depth is 9 ft.

True Statements

The swimmer comes back up 11 seconds after the timer is started.

The swimmer dives into the water 5 seconds after the timer is started.