Question

What is the vertex for the function? (You will want to find the axis of symmetry first.) Write as an ordered pair, leaving no spaces, such as (1,-4).
y = -3x2 + 6x

Answer 1

hmm for a quadratic equation like that \(\bf \Large ax^2+bx+c\) , which is just a parabola, you can find the vertex at the coordinates at \(\bf \left(-\cfrac{b}{2a}\quad ,\quad c-\cfrac{b^2}{4a}\right)\)

Answer 2

notiice that \(\bf y=-3x^2+6x \implies y=-3x^2+6x+0\)

Answer 3

Okay, I see that.

Answer 4

\(\bf y = -3x^2 + 6x+0\\ \quad \\
\left(-\cfrac{b}{2a}\quad ,\quad c-\cfrac{b^2}{4a}\right)\implies
\left(-\cfrac{(6)}{2(-3)}\quad ,\quad (0)-\cfrac{(6)^2}{4(-3)}\right)\)

Answer 5

Okay, I'm sorry I'm not too good in math. Cam you explain this a bit more for me?

Answer 6

well.... is simple.... just evaluate the fractions

Answer 7

okay, so (6).2(-3), and 6^2/4(-3)?

Answer 8

\(\large \begin{array}{llll}
y = -3x^2 + 6x+0\\ \quad \\
\left(-\cfrac{b}{2a}\quad ,\quad c-\cfrac{b^2}{4a}\right)\implies
&\left(-\cfrac{(6)}{2(-3)}\quad ,\quad (0)-\cfrac{(6)^2}{4(-3)}\right)\\
&\qquad x\qquad \qquad \qquad \qquad y

\end{array}\)

Answer 9

Okay, awesome, thanks!

Answer 10

yw