What is the vertex of $y= 3x^2-x+7(x-5)^2$ ?

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Answer 1 · 2018-05-06T09:19:14

Vertex is $(3 11/20,48 39/40)$

Let us convert this to vertex form $y=a(x-h)^2+k$ , which has $(h,k)$ as vertex

$y=3x^2-x+7(x-5)^2$

= $3x^2-x+7(x^2-10x+25)$

= $3x^2-x+7x^2-70x+175$

= $10x^2-71x+175$

= $10(x^2-71/10x)+175$

= $10(x^2-2*71/20*x+(71/20)^2-(71/20)^2)+175$

= $10(x^2-2*71/20*x+(71/20)^2)-10(71/20)^2+175$

= $10(x-71/20)^2+1959/40$

= $10(s-3 11/20)^2+48 39/40$

Hence vertex is $(3 11/20,48 39/40)$

What is the vertex of y= 3x^2-x+7(x-5)^2 y= 3x^2-x+7(x-5)^2 ?