Respuesta :
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 [tex]y=5x^2+10x-120[/tex] 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.
[tex]y=a(x+6)(x-4)[/tex] ...(i)
The parabola crosses the y- axis at -120. It means the equation of the parabola must be true for (0,-120).
[tex]-120=a(0+6)(0-4)[/tex]
[tex]-120=a(6)(-4)[/tex]
[tex]-120=-24a[/tex]
Divide both sides by -24.
[tex]\dfrac{-120}{-24}=a[/tex]
[tex]5=a[/tex]
Substituting [tex]a=5[/tex] in (i), we get
[tex]y=5(x+6)(x-4)[/tex]
[tex]y=5(x^2+6x-4x-24)[/tex]
[tex]y=5(x^2+2x-24)[/tex]
[tex]y=5x^2+10x-120[/tex]
So, the equation of the parabola is [tex]y=5x^2+10x-120[/tex].
The vertex of a parabola [tex]f(x)=ax^2+bx+c[/tex] is:
[tex]Vertex=\left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right)[/tex]
In the equation of the parabola, [tex]a=5,b=10,c=-120[/tex].
[tex]-\dfrac{b}{2a}=-\dfrac{10}{2(5)}[/tex]
[tex]-\dfrac{b}{2a}=-\dfrac{10}{10}[/tex]
[tex]-\dfrac{b}{2a}=-1[/tex]
Putting [tex]x=-1[/tex] in the equation of the parabola, we get
[tex]y=5(-1)^2+10(-1)-120[/tex]
[tex]y=5-10-120[/tex]
[tex]y=-125[/tex]
So, the vertex of the parabola is at point (-1,-125).
Therefore, Lucas is correct.
