• Not so easy. First have to assume the parabola is aligned with the y axis. Intuitively the answer is: parabolas are symmetrical. A line parallel to the y axis through the vertex is like a mirror. Therefore the two zeros are mirror images and their midpoint is on the mirror and has the same x-coordinate as the vertex. Mathematically ... Let the zeros be p and q Then the parabola is of the form a*(x-p)*(x-q) Let's change the variable so it is centered on the mid point of the zeros. Put x = t + mid point so that t is zero when x is on the mid point. x = t + ((p+q)/2) Now what does the parabola look like: a*(t+((p+q)/2)-p)*(t+((p-q)/2)-q) some rearrangement: a*(t-((p-q)/2)*(t+((p-q)/2)) which is a*(t-something)*(t+something) which is a*(t^2 - somthing^2) Note: t^2 is symmetrical: it has the same form if t is replaced with -t. All we have to do is find the t coordinate of the vertex. The vertex occurs at the extremum of the parabola, when it takes its greatest or least value. The expression a*(t^2 - something^2) takes an extremum when t^2 does. t^2 has an extremum, in this case a minimum, at t=0 So the x coordinate of the vertex is the same as the mid point of the zeros.

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