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

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Answer 1 · 2017-09-07T21:21:26

See below.

$4/3( x - 2)^2 -5$

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

Multiply bracket by $4/3$

$(4x^2)/3 - 4/3x + 4/3 - 5$

Simplify:

$4/3x^2 - 4/3x - 11/3$

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