# A missile is launched from the ground. Its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. After 1 second, the missile is 130 feet in the air; after 2 seconds, it is 240 feet in the air. Find the height, in feet, of the missile after 11 seconds in the air.

Maybe its too late for this one but, its better to give you the answer to your question. 1. Find slope using two points. If you think about it, the missle's height can be graphed, therefore its position at x can be used as points. We were already given two points, we just didn't realize they were points. To find slope: (Y2 - Y1) / (X2 - X1) where Point1 = (1, 130) and Point2 = (2,240) Therefore, (240 - 130) / (2 - 1) = 110 m/s This is the velocity since the slope (also called a derivative) of the position is just velocity 2. Use point-slope form to make an equation. Point-slope form says that if you are given a point and a slope you are GUARANTEED a function. We have our slope: 110 We have two points, we just pick one of them: (1,130) We know point-slope form: y - y1 = m(x - x1) So solve: y - 130 = 110(x - 1) y = 110x - 110 + 130 y = 110x + 20 This is now h(x) = 110x + 20. 3. We wanted the height at 11 seconds so we plug'n'chug 11 for x. h(11) = 110(11) + 20 h(11) = 1210 + 20 h(11) = 1230 feet That's it! RAWR!