Question

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

Answer 1

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

Explanation:

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)`