SOLUTION: 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 equatio

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 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 equatio      Log On


   

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
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
I did this quickly on paper, and have not rechecked this result yet:
%28x-1%2F2%29%5E2=-4%281%2F16%29%28y-1%29




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The first two given points show the obvious roots.
If Q is a nonzero factor,
y=Q%28x-0%29%28x-1%29
A few simple algebra steps using the last given point,
Q=-4

Continuing with y=-4x%28x-1%29
multiplying, and later completing the square,
.
.
%28x-1%2F2%29%5E2=-%281%2F4%29%28y-1%29
Or if you want the form to be as %28x-1%2F2%29%5E2=-4a%28x-1%29
then find that a=1%2F16

highlight_green%28%28x-1%2F2%29%5E2=-4%281%2F16%29%28x-1%29%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

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 
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%281%2C5%2C+h%2C+%22=%22%2C+%281+-+0%29%2F2%2C+%22=%22%2C+1%2F2%29

 matrix%281%2C3%2C+%28x+-+h%29%5E2%2C+%22=%22%2C+-+4a%28y+-+k%29%29
matrix%281%2C3%2C+%280+-+1%2F2%29%5E2%2C+%22=%22%2C+-+4a%280+-+k%29%29 ----- Substituting 1%2F2 for h, and (0, 0) for (x, y)
      matrix%281%2C3%2C+1%2F4%2C+%22=%22%2C+4ak%29


 matrix%281%2C3%2C+%28x+-+h%29%5E2%2C+%22=%22%2C+-+4a%28y+-+k%29%29
matrix%281%2C3%2C+%285+-+1%2F2%29%5E2%2C+%22=%22%2C+-+4a%28-+20+-+k%29%29 -- Substituting 1%2F2 for h, and (5, - 20) for (x, y)
 
     matrix%281%2C3%2C+81%2F4%2C+%22=%22%2C+80a+%2B+1%2F4%29 ----- Substituting 1%2F4 for 4ak
  

      matrix%281%2C3%2C+1%2F4%2C+%22=%22%2C+4ak%29 
      matrix%282%2C3%2C+1%2F4%2C+%22=%22%2C+4%281%2F4%29k%2C+1%2F4%2C+%22=%22%2C+k%29 ----- Substituting matrix%281%2C3%2C+1%2F4%2C+for%2C+a%29

  matrix%281%2C3%2C+%28x+-+h%29%5E2%2C+%22=%22%2C+-+4a%28y+-+k%29%29
 matrix%281%2C3%2C+%28x+-+1%2F2%29%5E2%2C+%22=%22%2C+-+4%281%2F4%29%28y+-+1%2F4%29%29 ------ Substituting 1%2F2 for h, matrix%281%2C3%2C+1%2F4%2C+for%2C+a%29, and 1%2F4 for k
  <===== Correct equation