How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=2x^2-4x+1#?

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Answer 1 · 2017-02-10T13:24:26

The y intercept is the point #(0;1)#

The axis of symmetry is #x=1#

The vertex is #V=(1;-1)#

The y intercept is the point #(0;1) (x=0->y=1)#

The axis of symmetry is calculated by the formula:

#x=-b/(2a)#

therefore, since #a=2# and #b=-4#, you get:

#x=-(-4)/(2*2)=1#

The vertex is the one point of the axis of symmetry that belongs to the function, then:

#x_V=1#

#y_V=f(1)=2*1^2-4*1+1=-1#

Then #V=(1;-1)#

graph{2x^2-4x+1 [-2, 5, -2, 5]}