Find an equation in standard form of the parabola passing through the points (1, -2), (2, -2), (3, -4)?

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  y = -x^2 +3x -4

Step-by-step explanation:

We are given two points with the same y-value, so we know the axis of symmetry lies halfway between their x-coordinates, at ...

  x = (1 +2)/2 = 3/2

Then the vertex form of the equation can be written ...

  y = a(x -3/2)^2 +k

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We can use two of the points to find the values of 'a' and 'k'.

For (x, y) = (1, -2)

 -2 = a(1 -3/2)^2 +k = a/4 +k

For (x, y) = (3, -4)

  -4 = a(3 -3/2)^2 +k = 9a/4 +k

Subtracting the first of these equations from the second, we have ...

  (9a/4 +k) -(a/4 +k) = (-4) -(-2)

  2a = -2

  a = -1

Then k can be found from the first equation.

  -2 = -1/4 +k

  k = -1  3/4

So, the vertex form equation is ...

  y = -(x -3/2)^2 -7/4

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Expanding this, we can find the standard form.

  y = -(x^2 -3x +9/4) -7/4)

  y = -x^2 +3x -4

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