Question 896300
since you're vertex is above the x-axis, then your quadratic equation will have to point up and open down if there are going to be any x-intercepts.


the vertex form of a quadratic equation is:


y = a * (x-h)^2 + k


the quadratic equation will point up and open down if the coefficient of the x^2 term have to be negative, so the general equation of the vert4ex form will be:


y = -a * (x-h)^2 + k


(h,k) is the vertex which is equal to (2,4) as given.


your equation becomes:


y - -a * (x-2)^2 + 4


a can be any value and your vertex will be at (2,4).


the problem is that the x-intercept will be different when the value of a is different.


the following graph shows what i mean.


<img src = "http://theo.x10hosting.com/2014/083002.jpg" alt="$$$" </>


bottom line is you don't have enough information to solve this problem if all that you are given is the vertex with no other information that can help you to narrow down  the solution.


as you stated, you don't even know if the parabole points up or down.


you don't even know if the parabola points left or right instead of up or down.


just knowing where the vertex is doesn't tell you anything other than where the vertex is.


you need more information that does not appear to be there.