What is the the vertex of $x = {(y - 6)}^{2} - y + 1$ $x = (y - 6)^2 - y+1$ ?

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Answer 1 · 2017-12-28T16:06:27

Vertex is $(- 5 \frac{1}{4}, - 6 \frac{1}{2})$ $(-5 1/4,-6 1/2)$

We can write $x = {(y - 6)}^{2} - y + 1$ $x=(y-6)^2-y+1$ as

$x = y^{2} - 12 y + 36 - y + 1$ $x=y^2-12y+36-y+1$

= $y^{2} - 13 y + {(\frac{13}{2})}^{2} - \frac{169}{4} + 37$ $y^2-13y+(13/2)^2-169/4+37$

= ${(y - \frac{13}{2})}^{2} - \frac{169 - 148}{4}$ $(y-13/2)^2-(169-148)/4$

= ${(y - \frac{13}{2})}^{2} - \frac{21}{4}$ $(y-13/2)^2-21/4$

Hence vertex is $(- \frac{21}{4}, - \frac{13}{2})$ $(-21/4,-13/2)$ or $(- 5 \frac{1}{4}, - 6 \frac{1}{2})$ $(-5 1/4,-6 1/2)$

What is the the vertex of x=(y−6)2−y+1x = (y - 6)^2 - y+1?