Question

Find the coordinates of the vertex of the parabola.
y = 9x^2 + 12x + 3

Answer 1

do you know how to complete the square?

Answer 2

\[ax^2+bx+c \\ ax^2+\frac{a}{a}bx+c \\ \text{ first step factor out the coefficient of } x^2 \\ \text{ from the first two terms } \\ a(x^2+\frac{1}{a}bx)+c\]
you try with yours.

Answer 3

then we can complete the square look at the number in front of x...
we are going to add in (b/(2a))^2 inside the ( )
but whatever we add in we have to subtract out
\[a(x^2+\frac{b}{a}x)+c \\ a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2-(\frac{b}{2a})^2)+c \\ a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)-a(\frac{b}{2a})^2 +c \text{ by distributive law } \\ a(x+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c \\ \text{ we have written the quadratic expression in vertex form now }\]

Answer 4

this is called vertex form because it tells us what the vertex is

Answer 5

\[(-\frac{b}{2a},c-a(\frac{b}{2a})^2)\]

Answer 6

I'm still confused