Question

The x- intercepts of a parabola are (0,-6) and (0,4). The parabola crosses the y- axis at -120. Lucas said that an equation for the parabola is y=5x^2+10x-120 and that the coordinates of the vertex are (-1, -125). Do you agree or disagree? List why?

Answer 1

Given:

The x- intercepts of a parabola are (0,-6) and (0,4).

The parabola crosses the y- axis at -120.

Lucas said that an equation for the parabola is
`y=5x^2+10x-120` and that the coordinates of the vertex are (-1, -125).

To find:

Whether Lucas is correct or not.

Solution:

The x- intercepts of a parabola are (0,-6) and (0,4). It means (x+6) and (x-4) are the factors of the equation of the parabola.

`y=a(x+6)(x-4)` ...(i)

The parabola crosses the y- axis at -120. It means the equation of the parabola must be true for (0,-120).

`-120=a(0+6)(0-4)`

`-120=a(6)(-4)`

`-120=-24a`

Divide both sides by -24.

`(-120)/(-24)=a`

`5=a`

Substituting
`a=5` in (i), we get

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

`y=5(x^2+6x-4x-24)`

`y=5(x^2+2x-24)`

`y=5x^2+10x-120`

So, the equation of the parabola is
`y=5x^2+10x-120`.

The vertex of a parabola
`f(x)=ax^2+bx+c` is:

`Vertex=\left(-(b)/(2a),f(-(b)/(2a))\right)`

In the equation of the parabola,
`a=5,b=10,c=-120`.

`-(b)/(2a)=-(10)/(2(5))`

`-(b)/(2a)=-(10)/(10)`

`-(b)/(2a)=-1`

Putting
`x=-1` in the equation of the parabola, we get

`y=5(-1)^2+10(-1)-120`

`y=5-10-120`

`y=-125`

So, the vertex of the parabola is at point (-1,-125).

Therefore, Lucas is correct.