Question

the vertex of y=2x^2+5x+4

Answer 1

Crud, I meant...
\[\Large \left(-\frac{\color{blue}b}{2\color{red}a}\quad , \quad \frac{4\color{red}a\color{green}c-\color{blue}b^2}{4\color{red}a}\right)\]

Answer 2

\[\Large y = \color{red}ax^2 + \color{blue}bx + \color{green}c\]

then the vertex is \[\Large \left(-\frac{\color{blue}b}{2\color{red}a}\quad , \quad \frac{4\color{red}a\color{green}c-\color{blue}b^2}{4\color{red}a}\right)\]

Answer 3

Just plug in and be on your way :D

Answer 4

Or you could do that D:

Answer 5

lol

Answer 6

better parentheses though :3

Answer 7

You know what...... im making a whole bunch of typos -_-

Answer 8

\begin{align}
y=2x^2 +5x+4 &= 2\left(x^2+\cfrac{5}{4}x\right)+4 \\
&=2\left(x^2 +\cfrac{5}{4}+\cfrac{25}{16}-\cfrac{25}{16}\right) +4 \\
&= 2\left(x+\cfrac{5}{4}\right)^2 +\cfrac{7}{8} \\
\end{align}

Vertex form: \(y= a(x-h)^2 +k\)
Your form: \(y=2\left(x+\cfrac{5}{4}\right)^2 +\cfrac{7}{8}\)

Answer 9

\[\Large \color{red}ax^2 +\color{blue}bx+\color{green}c\]\[\Large \color{red}a\left(x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{green}c}{\color{red}a}\right)\]\[\Large \color{red}a\left(x^2 +\frac{\color{blue}b}{\color{red}a}x+\frac{\color{blue}b^2}{4\color{red}a^2}-\frac{\color{blue}b^2}{4\color{red}a^2}+\frac{\color{green}c}{\color{red}a}\right)\]\[\Large \color{red}a\left[\left(x +\frac{\color{blue}b}{2\color{red}a}\right)^2-\frac{\color{blue}b^2}{4\color{red}a^2}+\frac{\color{green}c}{\color{red}a}\right]\]\[\Large \color{red}a\left[\left(x +\frac{\color{blue}b}{2\color{red}a}\right)^2+\frac{4\color{red}a\color{green}c-\color{blue}b^2}{4\color{red}a^2}\right]\]

\[\Large \color{red}a\left(x +\frac{\color{blue}b}{2\color{red}a}\right)^2+\frac{4\color{red}a\color{green}c-\color{blue}b^2}{4\color{red}a} \]

@Jhannybean what now? XD

Answer 10

=______= ooooook Terence!

Answer 11

Don't I just love being general? :D

Answer 12

lol...

Answer 13

^_^