Question 1187729

can I ask this for last time, it's just that the other answer is correct but will use another equation.

axis vertical and passing through (0,0),(1,0),(5,-20) 
using the equation: (x-h)^2=-4a(y-k)

thank you so much, sorry for the inconvenience 
<pre>The x-value that's MIDWAY between the 2 ROOTS is the x-coordinate of the vertex, or h. 
With the ROOTS being (0, 0) and (1, 0), {{{matrix(1,5, h, "=", (1 - 0)/2, "=", 1/2)}}}

 {{{matrix(1,3, (x - h)^2, "=", - 4a(y - k))}}}
{{{matrix(1,3, (0 - 1/2)^2, "=", - 4a(0 - k))}}} ----- Substituting {{{1/2}}} for h, and (0, 0) for (x, y)
      {{{matrix(1,3, 1/4, "=", 4ak)}}}


 {{{matrix(1,3, (x - h)^2, "=", - 4a(y - k))}}}
{{{matrix(1,3, (5 - 1/2)^2, "=", - 4a(- 20 - k))}}} -- Substituting {{{1/2}}} for h, and (5, - 20) for (x, y)
 {{{matrix(2,3, (4&1/2)^2, "=", 80a + 4ak, (9/2)^2, "=", 80a + 4ak)}}}
     {{{matrix(1,3, 81/4, "=", 80a + 1/4)}}} ----- Substituting {{{1/4}}} for 4ak
  {{{matrix(3,3, 81/4 - 1/4, "=", 80a, 80/4, "=", 80a, 1/4, "=", a)}}}

      {{{matrix(1,3, 1/4, "=", 4ak)}}} 
      {{{matrix(2,3, 1/4, "=", 4(1/4)k, 1/4, "=", k)}}} ----- Substituting {{{matrix(1,3, 1/4, for, a)}}}

  {{{matrix(1,3, (x - h)^2, "=", - 4a(y - k))}}}
 {{{matrix(1,3, (x - 1/2)^2, "=", - 4(1/4)(y - 1/4))}}} ------ Substituting {{{1/2}}} for h, {{{matrix(1,3, 1/4, for, a)}}}, and {{{1/4}}} for k
 {{{highlight_green(matrix(1,3, (x - 1/2)^2, "=", - (y - 1/4)))}}} <===== Correct equation</pre>