The standard form of a quadratic equation is [tex]\displaystyle{ y=ax^2+bx+c[/tex], while the vertex form is:
[tex]y=a(x-h)^2+k[/tex], where (h, k) is the vertex of the parabola.
What we want is to write [tex]\displaystyle{ y=3x^2-18x-6[/tex] as [tex]y=a(x-h)^2+k[/tex]
First, we note that all the three terms have a factor of 3, so we factorize it and write:
[tex]\displaystyle{ y=3(x^2-6x-2)[/tex].
Second, we notice that [tex]x^2-6x[/tex] are the terms produced by [tex](x-3)^2=x^2-6x+9[/tex], without the 9. So we can write:
[tex]x^2-6x=(x-3)^2-9[/tex], and substituting in [tex]\displaystyle{ y=3(x^2-6x-2)[/tex] we have:
[tex]\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11][/tex].
Finally, distributing 3 over the two terms in the brackets we have:
[tex]y=3[x-3]^2-33[/tex].
Answer: [tex]y=3(x-3)^2-33[/tex]
