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Find the equation of the parabola in standard form which passes through the points $(-2,1)$, $(1,2)$ and $(-1,3)$ with axis parallel to the x-axis.

Hint: Use the form $y^2 + Dx + Ey + F = 0$.

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    It would be nice for you to tell others what progress you've made. Also where did you encounter a problem in your try? – 2017-01-23

2 Answers 2

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HINT: Plug the three points in the equation and you will get three linear equations for three unknowns, which should be easily solved.

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You get this inhomogeneous linear system in $D, E, F$: $$ A x = b \iff \begin{pmatrix} -2 & 1 & 1 \\ 1 & 2 & 1 \\ -1 & 3 & 1 \end{pmatrix} \begin{pmatrix} D \\ E \\ F \end{pmatrix} = \begin{pmatrix} -1 \\ -4 \\ -9 \end{pmatrix} $$

Solving in Octave:

>> A = [-2,1,1; 1,2,1;-1,3,1]
A =

  -2   1   1
   1   2   1
  -1   3   1

>> b = [-1;-4;-9]
b =

  -1
  -4
  -9

>> inv(A)
ans =

  -0.20000   0.40000  -0.20000
  -0.40000  -0.20000   0.60000
   1.00000   1.00000  -1.00000

>> x = inv(A) * b
x =

   0.40000
  -4.20000
   4.00000

Asking my friend Ruby, if those parameters work:

irb> def f(x,y) y**2 + 0.4 * x + -4.2 * y + 4 end
=> :f
irb> f(-2,1)
=> 0.0
irb> f(1,2)
=> 0.0
irb> f(-1,3)
=> -1.7763568394002505e-15