Given:
The eyuation of the parabola.
[tex]y=(x-4)(x+2)[/tex]Required:
We need to find the x-intercepts, vertex, and standard form of the equation.
Explanation:
Set y =0 and solve for x to find the x-intercepts of the parabola.
[tex](x-4)(x+2)=0[/tex][tex](x-4)=0,(x+2)=0[/tex][tex]x=4,x=-2[/tex]The x-intercepts are 4 and -2.
Multipy (x-4) and (x+2) to find the stansdad form of the equation.
[tex]y=x\left(x+2\right)-4\left(x+2\right)[/tex][tex]y=(x)x+2(x)+(-4)x+(-4)2[/tex][tex]y=x^2+2x-4x-8[/tex][tex]y=x^2-2x-8[/tex]The standard form of the equation is
[tex]y=x^2-2x-8.[/tex]which is of the fom
[tex]y=ax^2+bx+c[/tex]where a =1, b =-2 and c =-8.
[tex]\text{ The x- coordinate of the vertex is }h=-\frac{b}{2a}.[/tex]Substitute b =-2 and a =1 in the equation.
[tex]\text{ The x- coordinate of the vertex is }h=-\frac{(-2)}{2(1)}=1[/tex][tex]substitute\text{ x =1 in the equation }y=x^2-2x-8\text{ to find the y-coordinate of the vertex.}[/tex][tex]y=1^2-2(1)-8=-9[/tex]The vertex of the given parabola is (1,-9).
Final answer:
1)
The x-intercepts are 4 and -2.
2)
The standard form of the equation is
[tex]y=x^2-2x-8.[/tex]3)
The vertex of the given parabola is (1,-9).
