The maximum value can be determined by taking the derivative of the function.
(dh/dt) [h(t)] = h'(x) = -9.8t + 6
Set h'(x) = 0 to find the critical point
-9.8t + 6 = 0
-9.8t = -6
t = 6/9.8
Plug the time back into the function to find the height.
h(6/9.8) = -4.9(6/9.8)^2 + 6(6/9.8) + .6
= 2.4
And I don't understand your second question.
So -2 is the domain so it is 2 left.
Then -3 is the range so you go 3 down.
so line M is the answer
<span>3x = y + 7
</span><span> y =3x-7
</span>
<span>8x = 2y + 5
</span><span>8x = 2(3x-7) + 5
8x=6x-14+5
8x-6x=-9
2x=-9
x=-4.5
</span><span>8x = 2y + 5
</span>8(4.5)=2y+5
36=2y+5
2y=31
y=15.5
m<3 =72°
This is because m<1 and m<3 are vertically opposite.
y= -4x² - 16x - 14
Take the coefficient of x² as the common factor
y = - 4(x² + 4x + 7/2)
Add (b/2)² to complete the square
y = - 4( x² + 4x + 2² - 2² + 7/2)
Complete the square
y = -4( (x + 2)² - 1/2)
Remove the bracket
y = -4(x + 2)² + 2
Answer: y = -4(x + 2)² + 2
