What is the line of symmetry for the parabola whose equation is #y=2x^2-4x+1#?

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Answer 1 · 2018-06-12T17:39:36

#x=1#

Method 1: Calculus approach.

#y=2x^{2}-4x+1#
#\frac{dy}{dx}=4x-4#
The line of symmetry will be where the curve turns (due to the nature of the #x^{2}# graph.

This is also when the gradient of the curve is 0.

Therefore, let #\frac{dy}{dx}=0#

This forms an equation such that:
#4x-4=0#
solve for x, #x=1# and line of symmetry falls on the line #x=1#

Method 2: Algebraic approach.

Complete the square to find the turning points:

#y=2(x^2-2x+\frac{1}{2})#
#y=2((x-1)^{2}-1+\frac{1}{2})#
#y=2(x-1)^{2}-1#

From this we can pick up the line of symmetry such that:
#x=1#