The equation of a parabola with vertex (h,k) is:
[tex]f(x)=a(x-h)^2+k[/tex]
Where a is a constant.
Rewrite the given expression in vertex form by completing the square:
[tex]\begin{gathered} f(x)=2x^2-2x+1 \\ =2(x^2-x)+1 \\ =2(x^2-x+\frac{1}{4}-\frac{1}{4})+1 \\ =2(x^2-x+\frac{1}{4})-2\cdot\frac{1}{4}+1 \\ =2(x-\frac{1}{2})^2-\frac{1}{2}+1 \\ =2(x-\frac{1}{2})^2+\frac{1}{2} \end{gathered}[/tex]
Use the expression in vertex form to answer parts a to c.
a)
To find the y-intercept, evaluate f at x=0:
[tex]f(0)=2(0)^2-2(0)+1=1[/tex]
The axis of symmetry is a vertical line that passes through the vertex. Since the coordinates of the vertex are (1/2,1/2), then the equation of the axis of symmetry is:
[tex]x=\frac{1}{2}[/tex]
And the x-coordinate of the vertex is 1/2.
b)
Use the values x=0, x=0.5, x=1, x=1.5, x=2 to make a table:
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