Question

What is the equation of the parabola that has a vertex at `(2, 5)` and passes through point `(1,-1)`?

Answer 1

`y=-6x^2+24x-19` the standard form

`(x-2)^2=-1/6(y-5)` the vertex form

Explanation:

Assume the parabola opening downward because, the additional point is below the Vertex

Given Vertex at `(2, 5)` and passing through `(1, -1)`

Solve for `p` first

Using Vertex form `(x-h)^2=-4p(y-k)`

`(1-2)^2=-4p(-1-5)`

`(-1)^2=-4p(-6)`

`1=24p`

`p=1/24`

Use now Vertex form `(x-h)^2=-4p(y-k)` again with variables x and y only

`(x-2)^2=-4(1/24)(y-5)`

`(x-2)^2=-1/6(y-5)`

`-6(x^2-4x+4)+5=y`

`y=-6x^2+24x-24+5`

`y=-6x^2+24x-19`

kindly check the graph

graph{y=-6x^2+24x-19[-25,25,-12,12]}

Answer 2

The equation of paqrabola is `y= -6*x^2+24*x-19`

Explanation:

The equation o0f the parabola is `y=a*(x-h)^2+k` Where (h,k) is the co-ordinates of vertex. So `y = a*(x-2)^2 +5` Now the Parabola passes through point(1,-1) so `-1 = a*(1-2)^2+5 or -1=a+5 or a=-6`
Now putting the value of a in the equation of parabola we get `y=-6(x-2)^2+5 or y= -6*x^2+24*x-19`
graph{-6x^2+24x-19 [-10, 10, -5, 5]} [Answer]