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