What is the vertex of $y = x - 2 + {(x - 3)}^{2}$ $y=x-2 +(x-3)^2$ ?

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Answer 1 · 2017-11-14T11:43:02

Vertex is at $(2.5, 0.75)$ $(2.5,0.75)$

$y = x - 2 + {(x - 3)}^{2} or y = x - 2 + x^{2} - 6 x + 9$ $y=x-2+(x-3)^2 or y=x-2+x^2-6x+9$ or

$y = x^{2} - 5 x + 7 or y = (x^{2} - 5 x) + 7$ $y=x^2-5x+7 or y=(x^2-5x) +7$ or

$y = {x^{2} - 5 x + {(\frac{5}{2})}^{2}} - \frac{25}{4} + 7$ $y={x^2-5x +(5/2)^2}-25/4 +7$ or

$y = {(x - 2.5)}^{2} + \frac{3}{4} or y = {{x - 2.5)}^{2} + 0.75$ $y=(x-2.5)^2+3/4 or y={x-2.5)^2+0.75$

Comparing with vertex form of equation

$y = a {(x - h)}^{2} + k; (h, k)$ $y = a(x-h)^2+k ; (h,k)$ being vertex we find

here $h=2.5 , k=0.75 :.$ Vertex is at $(2.5,0.75)$ .

graph{(x-2)+(x-3)^2 [-10, 10, -5, 5]} [Ans]

What is the vertex of y=x-2 +(x-3)^2y=x−2+(x−3)2y=x-2 +(x-3)^2?