# What is m if vertex of parabola y=3(m+1)x^2+6mx+m+3 is on line y=x-3?

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### 1 Answer

You need to remember what coordinates of vertex of parabola are such that:

`x = -(6m)/(2*3(m+1))`

`x = -m/(m+1)`

`y = (4*3(m+1)(m+3) - (6m)^2)/(4*3(m+1))`

`y = (12m^2 + 48m + 36 - 36m^2)/(12(m+1))`

`y = (-24m^2 + 48m + 36)/(12(m+1))`

You should remember that a point lies on a line if its coordinates check the equation of the line, hence, you need to substitute x and y of vertex of parabola in equation of line y = x-3 such that:

`(-24m^2 + 48m + 36)/(12(m+1)) = -m/(m+1) - 3`

You need to bring the terms from the right to a common denominator such that:

`-24m^2 + 48m + 36 = -12m - 36(m+1)`

You need to bring all terms to the left side such that:

`-24m^2 + 48m + 36 + 12m + 36m + 36 = 0`

`-24m^2 + 96m + 72 = 0`

You need to divide by -4 such that:

`6m^2- 24m-18 = 0`

You need to divide by 6 such that:

`m^2 - 4m - 3 = 0`

You should use quadratic formula such that:

`m_(1,2) = (4+-sqrt(16+12))/2`

`m_(1,2) = (4+-sqrt28)/2`

`m_(1,2) = (4+-2sqrt7)/2`

`m_(1,2) = (2+-sqrt7)`

**Hence, evaluating the values of m for the vertex of parabola to lie on the given line yields `m_(1,2) = (2+-sqrt7).` **

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