Help if you can do this please

Step-by-step explanation:

a. Since the parabola is compressed by a factor of 1/3 we can state:

a parabola is written this way : y=(x-h)²+k h stands for the translation to the left ⇒ 2*3=6k for the units down  ⇒4*3=12

So the equation is : y=(x-6)²+12

b.Here the parabola is stretched by a factor of 2 so we must multiply by 1/2

We khow that a parabola is written this way : y=(x-h)²+k (h,k) are the coordinates of the vertex the maximum value is  7*0.5=3.5 we khow tha the derivative of a quadratic function is null in the maximum value so let's derivate (x-h)²+k= x²+h²-2xh+k f'(x)= 2x-2h    h is 1 since the axe of simmetry is x=1 f'(x)=2x-2 ⇒2x-2=0⇒ x= 1 Now we khow that 1 is the point where the derivative is null f(1)=3.53.5=(x-1)²+k 3.5= (1-1)²+k⇒ k=3.5

So the equation is : y=(x-1)²+3.5

7.

the maximum height is where the derivative equals 0

h= -5.25(t-4)²+86h= -5.25(t²-8t+16)+86h=-5.25t²+42t-84+86h=-5.25t²+42t+2

Let's derivate it :

f(x)= -10.5t+42 -10.5t+42=042=10.5tt= 42/10.5=4

When the height was at max t=4s

h(max)= -5.25(4-4)²+86 = 86 m

h was 86m