Can anybody help me answer a question based on a parabola equation?

Question

The problem is "The height (y) of Marla's model rocket can be represented by the equation y = -16x^2 + 70x + 37 where(x) represents the time in seconds since the rocket was launched, and height is expressed in feet. What was the maximum height of Marla's rocket to the nearest tenth of a foot?" I've got a lot of questions like that one and I have no idea how to answer them. Please help.

Answer 1

The maximum or minimum of a parabola occurs at the vertex. since this graph opens down (because of the negative on the 16x^2) this one has a maximum at thevertex.
You find the vertex by using the formula x = -b/2a
Your a = -16 and b = 70
x = -70/(2(-16)) = 70/32=35/16
Put this back into the equation for x and solve for y
y = -16(35/16)^2 + 70(35/16) + 37
y = 113.6

Max height to nearest tenth = 113.6 feet and it happens at 35/16 seconds (2.1875sec)

Answer 2

y = -16x^2 + 70x + 37

You just have to find the maxima of the equation. Take it's first derivative and set it to Zero. Then solve for x. Then substitute the x you found into the y equation again to solve for the maximum height.

In your example:
y' = -32x + 70 = 0
Therefore x = 70/32 = 2.1875 seconds

At x =2.1875
the height of the rocket reaches,
y = -16(2.1875)^2 + 70(2.1875) + 37
y = 113.5625 feet, or rounded to 113.5 feet
(** Edit - sorry I messed up the on the arithmetic **)

To check if this indeed is the maxima (not minima),
take the 2nd derivative,
y" = -32 which is < 0, meaning the slope of the curve is going downward or starting to decrease, i.e. it is the maxima.

Answer 3

The parabola has a vertical axis of symmetry, and on that axis it reaches its maximum height halfway between equal y values. At y = 0 it's on the ground at launch (if launched from the ground) and at impact. Since your launch height is 37, solve

37 = -16x² + 70x + 37
0 = -16x² + 70x
0 = 2x(-8x + 35)
2x = 0, x = 0 (launch)
-8x + 35 = 0
8x = 35
x = 4.375 (back at 37 ft).
halfway between is 2.1875, and at that time the rocket is
-16(2.1875)² + 70(2.1875) + 37 = 113.5625 ft high

calculus makes all that easier. derivative is -32x + 70, set it = 0 to solve for x.