Answer:
The rocket will reach its maximum height after 6.13 seconds
Step-by-step explanation:
To find the time of the maximum height of the rocket differentiate the equation of the height with respect to the time and then equate the differentiation by 0 to find the time of the maximum height
∵ y is the height of the rocket after launch, x seconds
∵ y = -16x² + 196x + 126
- Differentiate y with respect to x
∴ y' = -16(2)x + 196
∴ y' = -32x + 196
- Equate y' by 0
∴ 0 = -32x + 196
- Add 32x to both sides
∴ 32x = 196
- Divide both sides by 32
∴ x = 6.125 seconds
- Round it to the nearest hundredth
∴ x = 6.13 seconds
∴ The rocket will reach its maximum height after 6.13 seconds
There is another solution you can find the vertex point (h , k) of the graph of the quadratic equation y = ax² + bx + c, where h =
and k is the value of y at x = h and k is the maximum/minimum value
∵ a = -16 , b = 196
∴ 
∴ h = 6.125
∵ h is the value of x at the maximum height
∴ x = 6.125 seconds
- Round it to the nearest hundredth
∴ x = 6.13 seconds
Answer:
(3,2)
Step-by-step explanation:
when reflecting across the x axis the y value becomes - while the x value stays the same
Answer:
C. f(x) = x - 7 all over 4
Step-by-step explanation:
NB: Let f(x) = y
Exchange X and Y
Make y the subject
f(x) = 4x + 7
y = 4x + 7
x = 4y + 7
x - 7 = 4y
x - 7 all over 4 = 4 ÷ 4
y = x - 7 all over 4
Answer:
y=a(x-p)(x-q)
y=a(x+2+√2)(x+2-√2)
passing through point (-1,1)
substitute
1=a(-1+2+√2)(-1+2-√2)
1=a(1+√2)(1-√2)
1=a(1-2)
1=a(-1)
a=1/(-1)
a=-1
y=-(x+[2+√2])(x+[2-√2])
y=-(x2+4x+2)
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