What is the standard form of $y = {(\frac{7}{5} x - \frac{4}{7})}^{2} + 4$ $y= (7/5x-4/7)^2+4$ ?

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What is the standard form of $y = {(\frac{7}{5} x - \frac{4}{7})}^{2} + 4$ $y= (7/5x-4/7)^2+4$ ?

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Answer 1 · 2017-10-02T01:55:31

$y = \frac{49}{25} x^{2} - \frac{8}{5} x + \frac{212}{49}$ $y=49/25x^2 -8/5x + 212/49$

Explanation:

Basically you just expand out the bracket.

Rule for squaring things: the first squared, plus the last squared, plus twice the product of the two. (like if you had ${(x + 3)}^{2}$ $(x+3)^2$ it'd be $x^{2} + 3^{2} + twice (3 \cdot x) = x^{2} + 6 x + 9$ $x^2 + 3^2 + "twice" (3*x) = x^2+6x+9$ )

So, ${(\frac{7}{5} x - \frac{4}{7})}^{2}$ $(7/5x-4/7)^2$ will be ${(\frac{7}{5} x)}^{2}$ $(7/5x)^2$ + ${(- \frac{4}{7})}^{2}$ $(-4/7)^2$ + $2 (\frac{7}{5} x \cdot - \frac{4}{7})$ $2(7/5x*-4/7)$
$= \frac{49}{25} x^{2} + \frac{16}{49} - \frac{8}{5} x$ $=49/25x^2 + 16/49 -8/5x$

Now add the +4:

$= \frac{49}{25} x^{2} - \frac{8}{5} x + 4 + \frac{16}{49} = \frac{49}{25} x^{2} - \frac{8}{5} x + \frac{212}{49}$ $=49/25x^2 -8/5x + 4 + 16/49 = 49/25x^2 -8/5x +212/49$

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What is the standard form of $y = {(\frac{7}{5} x - \frac{4}{7})}^{2} + 4$ $y= (7/5x-4/7)^2+4$ ?

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