How do you write $y = \frac{2}{3} x^{2} - \frac{4}{5} x$ $y=2/3x^2-4/5x$ in vertex form?

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Answer 1 · 2018-02-24T11:02:45

$y = \frac{2}{3} {(x - \frac{3}{5})}^{2} - \frac{6}{25}$ $y=2/3(x-3/5)^2-6/25$

$y = \frac{2}{3} x^{2} - \frac{4}{5} x$ $y = 2/3 x^2 - 4/5 x$

Take out a factor of $\frac{2}{3}$ $2/3$ ,

$y = \frac{2}{3} [x^{2} - \frac{6}{5} x]$ $y = 2/3[x^2 - 6/5 x]$

Complete the square,

$y = \frac{2}{3} [{(x - \frac{3}{5})}^{2} - \frac{9}{25}]$ $y= 2/3[(x-3/5)^2-9/25]$

And multiply out,

$y = \frac{2}{3} {(x - \frac{3}{5})}^{2} - \frac{6}{25}$ $y=2/3(x-3/5)^2-6/25$

How do you write y=2/3x^2-4/5xy=23x2−45xy=2/3x^2-4/5x in vertex form?