Question 1097695
the vertex form of a quadratic equation is y = a * (x-h)^2 + k


(h,k) is the vertex of the equation.


you are given that the maximum height is 12 and that the zero crossing points are  x = 0 and x = 12.


this makes the vertex of the quadratic equation equal to (6,12).


in the equation of y = a * (x-h)^2 + k, replace h with 6 and k with 12 to get:


y = a * (x-6)^2 + 12


the value of y is 0 when x is equal to 0 or x is equal to 12.


when x = 0, you get:


0 = a * (0-6)^2 + 12 which becomes 0 = a * (-6)^2 + 12 which becomes 0 = 36a + 12.


solve for a to get a = -1/3.


when x = 12, you get:


0 = a * (12-6)^2 + 12 which becomes 0 = a * (6)^2 + 12 which becomes 0 = 36a + 12.


solve fora to get a = -1/3.


the vertex form of the equation becomes y = -1/3 * (x-6)^2 + 12.


the graph of that equation is shown below:


<img src = "http://theo.x10hosting.com/2017/101601.jpg" alt="$$$" >


you can convert this vertex form of the equation into the standard form of the equation by simplifying it.


start with y = -1/3 * (x-6)^2 + 12
simplify to get y = -1/3 * (x^2 - 12x + 36) + 12
simplify further to get y = -1/3 * x^2 + 4x - 12 + 12
simplify further to get y = -1/3 * x^2 + 4x


that's the standard form of the equation.


the graph of that equation is shown below:


<img src = "http://theo.x10hosting.com/2017/101602.jpg" alt="$$$" >