we know that
The equation of the vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex
The axis of symmetry is equal to the x-coordinate of the vertex
so
[tex]x=h[/tex] ------> axis of symmetry of a vertical parabola
we will determine in each case the axis of symmetry to determine the solution
case A) [tex]f(x)=2x^{2}+x-1[/tex]
Convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)+1=2x^{2}+x[/tex]
Factor the leading coefficient
[tex]f(x)+1=2(x^{2}+0.5x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)+1+0.125=2(x^{2}+0.5x+0.0625)[/tex]
[tex]f(x)+1.125=2(x^{2}+0.5x+0.0625)[/tex]
Rewrite as perfect squares
[tex]f(x)+1.125=2(x+0.25)^{2}[/tex]
[tex]f(x)=2(x+0.25)^{2}-1.125[/tex]
the vertex is the point [tex](-0.25,-1.125)[/tex]
the axis of symmetry is
[tex]x=-0.25=-\frac{1}{4}[/tex]
therefore
the function [tex]f(x)=2x^{2}+x-1[/tex] has an axis of symmetry at [tex]x=-\frac{1}{4}[/tex]
case B) [tex]f(x)=2x^{2}-x+1[/tex]
Convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)-1=2x^{2}-x[/tex]
Factor the leading coefficient
[tex]f(x)-1=2(x^{2}-0.5x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)-1+0.125=2(x^{2}-0.5x+0.0625)[/tex]
[tex]f(x)-0.875=2(x^{2}-0.5x+0.0625)[/tex]
Rewrite as perfect squares
[tex]f(x)-0.875=2(x-0.25)^{2}[/tex]
[tex]f(x)=2(x-0.25)^{2}+0.875[/tex]
the vertex is the point [tex](0.25,0.875)[/tex]
the axis of symmetry is
[tex]x=0.25=\frac{1}{4}[/tex]
therefore
the function [tex]f(x)=2x^{2}-x+1[/tex] does not have a symmetry axis in [tex]x=-\frac{1}{4}[/tex]
case C) [tex]f(x)=x^{2}+2x-1[/tex]
Convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)+1=x^{2}+2x[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)+1+1=x^{2}+2x+1[/tex]
[tex]f(x)+2=x^{2}+2x+1[/tex]
Rewrite as perfect squares
[tex]f(x)+2=(x+1)^{2}[/tex]
[tex]f(x)=(x+1)^{2}-2[/tex]
the vertex is the point [tex](-1,-2)[/tex]
the axis of symmetry is
[tex]x=-1[/tex]
therefore
the function [tex]f(x)=x^{2}+2x-1[/tex] does not have a symmetry axis in [tex]x=-\frac{1}{4}[/tex]
case D) [tex]f(x)=x^{2}-2x+1[/tex]
Convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)-1=x^{2}-2x[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)-1+1=x^{2}-2x+1[/tex]
[tex]f(x)=x^{2}-2x+1[/tex]
Rewrite as perfect squares
[tex]f(x)=(x-1)^{2}[/tex]
the vertex is the point [tex](1,0)[/tex]
the axis of symmetry is
[tex]x=1[/tex]
therefore
the function [tex]f(x)=x^{2}-2x+1[/tex] does not have a symmetry axis in [tex]x=-\frac{1}{4}[/tex]
the answer is
[tex]f(x)=2x^{2}+x-1[/tex]
