What is the vertex form of $y = (x - 5) {(x + 6)}^{2} - {(x - 5)}^{2} (x + 6)$ $y= (x-5)(x+6)^2 - (x-5)^2 (x+6)$ ?

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Answer 1 · 2016-01-12T17:07:08

y= $11 {(x + \frac{1}{2})}^{2} - \frac{1331}{4}$ $11(x+1/2)^2 - 1331/4$

On simplification, the given equation would be y= (x-5)(x+6){x+6-(x-5)}.

= $11 (x^{2} + x - 30)$ $11(x^2 +x-30)$ . This can be converted to vertex form by completing the square.

= $11 (x^{2} + x + \frac{1}{4} - \frac{1}{4} - 30)$ $11(x^2 +x +1/4 -1/4 -30)$

= $11 {(x + \frac{1}{2})}^{2} - \frac{1331}{4}$ $11(x+1/2)^2 - 1331/4$

What is the vertex form of y=(x−5)(x+6)2−(x−5)2(x+6)y= (x-5)(x+6)^2 - (x-5)^2 (x+6) ?