# find the equation of the parabola with vertex at (2,-1), axis parallel to the y-axis and passing through (3,-2).pls . please show me the solutions .thanks

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### 1 Answer

Since the axis of the parabola is parallel to the y-axis, it indicates that the parabola either opens upward or downward.

For parabola that opens upward/downward, the standard equation is:

`(x-h)^2 = 4a(y-k)`

where (h,k) - is the vertex

(x,y) - points along the parabola and

a - the distance between the focus and vertex

If 4a is positive, the parabola opens upward. If negative, the parabola opens downward.

To determine the direction of the parabola, substitute the values of vertex(2,-1) and point along the parabola (3,-2) to the above formula.

`(x-h)^2 = 4a (y-k)`

`(3-2)^2 = 4a (-2-(-1))`

` 1^2 = 4a (-1)`

`1 = -4a`

`-1 = 4a`

So, the parabola opens downward. Then, substitute the value of 4a to the formula and also the vertex to set-up the equation of the above problem.

`(x-h)^2 = 4a(y-k)`

`(x-2)^2 = -1(y-(-1))`

(x-2)^2 = -(y+1)

This is the standard equation of the parabola. This can be simplifed by expanding left side and distribute -1 to (y+1) .

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

x^2-2x-2x+4 = -y - 1

x^2-4x+4 = -y - 1

x^2-4x+5 = -y

y = -x^2+4x-5

And this is the general equation of the parabola.

Note that standard and general form are two ways to express the equation of parabola. They are the same.

**Hence, the equation of the parabola in standard form is `(x-2)^2 = -(y+1)` and in general form, it is y=-x^2+4x-5.**

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