• anonymous
A parent takes their child to a water slide which is shaped similar to a portion of a parabola (see diagram). There is an elevator 3m from the edge of the pool. The elevator takes the rider to the top of the slide which is 19m above ground. The rider slides down and falls into pool from height of 1m. The parent (he is 2m tall) is standing 1m from the elevator and wishes to take a picture of the child when they are closest to the parent. What is the MIN. distance between parent and child? Let the elevator be at (0,0) So the parent is at (1,2) The edge of the pool is (3,1) So, the edge of the p
  • Stacey Warren - Expert
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  • jamiebookeater
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  • anonymous
So, the edge of the pool is the vertex; (3,1) We know (0,19) Equation(s): f(x) = 2(x-3)^2 + 1 <-- vertex form f(x) = 2x^2 - 12x + 19 <-- standard form f'(x) = 4x - 12 Now I am 100% sure I got these equations correct. I am now stuck because I don't how to get that point (x,f(x)) that will give the minimum distance. Would it work if I compute the distance from this line equation's normal to the point using the equation: d = |Ax + By + C| / root[A^2 + B^2] Subbing in for (x,y), the point (1,2)? Back to top View user's profile Send private message

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