Answer:
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=6
- k 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)=3.5
- 3.5=(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+86
- h= -5.25(t²-8t+16)+86
- h=-5.25t²+42t-84+86
- h=-5.25t²+42t+2
Let's derivate it :
- f(x)= -10.5t+42
- -10.5t+42=0
- 42=10.5t
- t= 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
The answer to the question
I think that would be a rhombus.
The answer above is correct
