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Help if you can do this please

Help if you can do this please-example-1
User Pfc
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Answer:

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

User Damb
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