What is the vertex of #y= 3(x - 1)^2 + 1#?

1 Answer
Meave60
Jan 26, 2016

The vertex for this parabola is #(1,1)#.

Explanation:

This is an equation for a parabola. #y=3(x-1)^2+1# is in vertex form, #y=a(x-h)^2+k#, where #a=3, x=h, k=1#

The vertex is the minimum or maximum point on the parabola. In this case the vertex is the minimum point because #a# is greater than one, so the parabola opens upward.

The axis of symmetry is the vertical line, #x#, that divides the parabola into two equal halves. In vertex form, the axis of symmetry is designated as #(x=h)#, so #(x=1)#.

The vertex point in vertex form is #(h,k)#, which is #(1,1)#.

Resource: http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php

graph{y=3(x-1)^2+1 [-10, 10, -5, 5]}

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