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

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Answer 1 · 2017-09-07T21:21:26

See below.

$\frac{4}{3} {(x - 2)}^{2} - 5$ $4/3( x - 2)^2 -5$

Square the bracket first. $\frac{4}{3} (x^{2} - 4 x + 4) - 5$ $4/3( x^2 - 4x + 4 ) - 5$

Multiply bracket by $\frac{4}{3}$ $4/3$

$\frac{4 x^{2}}{3} - \frac{4}{3} x + \frac{4}{3} - 5$ $(4x^2)/3 - 4/3x + 4/3 - 5$

Simplify:

$\frac{4}{3} x^{2} - \frac{4}{3} x - \frac{11}{3}$ $4/3x^2 - 4/3x - 11/3$

How do you write y=43(x−2)2−5y=4/3(x-2)^2-5 in standard form?