# Model Airplane math questions?

myself and my student tutor are trying to figure out this question here that's got us baffled as to what to do next:

"The path of a model airplane in a dive is described by the relationship y=x^2-12x+40, where y is the height of the plane in metres and x is the time in seconds"

a) What was the minimum height of the plane?

b) At what height did the plane start the dive?

c) How long did it take to reach the same height in part b)?

d) How long did it take the plane to reach 200m?

Can anyone help me figure this out?

### 2 Answers

- Some BodyLv 71 decade agoFavorite Answer
a)

The shape is a parabola, so it's minimum height would be at the vertex. The x value of the vertex is found using

x = -b/2a

x = -(-12)/(2*1) = 6 sec

y(6 sec) = 6^2 - 12(6) + 40

y = 4 m

b)

Initial height is when t = 0.

y(0 sec) = 0^2 - 12(0) + 40

y = 40 m

c)

40 = x^2 - 12x + 40

0 = x^2 - 12x

0 = x(x-12)

x = 0 or x = 12

x = 0 is the initial time, so x=12 is the answer we're looking for.

12 seconds.

d)

200 = x^2 - 12x + 40

0 = x^2 - 12x - 160

0 = (x-20)(x+8)

x = -8, 20

Since time can't be negative, t=20 is the right answer.

20 seconds.

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- 1 decade ago
a. y=(x-6)^2+4 .Then minimum height

occurs for x=6 sec -->y=4

b. for x=0 sec -->y=40 m

c. 40=(x-6)^2+4 solving for x

we have (x-6)^2=36 -->x=12 sec

d. 200=(x-6)^2+4 -->(x-6)^2=196

(x-6)=14 or x-6=-14 (not acceptable)

x=20 sec

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