What is the standard form of $y = (x + 7) (x + 1) - {(3 x - 7)}^{2}$ $y= (x+7)(x+1)-(3x-7)^2$ ?

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Answer 1 · 2016-03-28T06:30:10

$y = - 8 x^{2} + 50 x - 42$ $y=-8x^2+50x-42$

From the given equation
$y = (x + 7) (x + 1) - {(3 x - 7)}^{2}$ $y=(x+7)(x+1)-(3x-7)^2$

We start by expanding the right side using multiplication

$y = x^{2} + 8 x + 7 - (9 x^{2} - 42 x + 49)$ $y=x^2+8x+7-(9x^2-42x+49)$

$y = x^{2} + 8 x + 7 - 9 x^{2} + 42 x - 49$ $y=x^2+8x+7-9x^2+42x-49$

Simplify

$y = - 8 x^{2} + 50 x - 42$ $y=-8x^2+50x-42$

graph{y=(x+7)(x+1)-(3x-7)^2[-80,80,-40,40]}

God bless....I hope the explanation is useful.

What is the standard form of y= (x+7)(x+1)-(3x-7)^2y=(x+7)(x+1)−(3x−7)2 y= (x+7)(x+1)-(3x-7)^2?