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.

$\dfrac{-120}{-24}=a$

$5=a$

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:

$Vertex=\left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right)$

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

$-\dfrac{b}{2a}=-\dfrac{10}{2(5)}$

$-\dfrac{b}{2a}=-\dfrac{10}{10}$

$-\dfrac{b}{2a}=-1$

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.