Answer: The required line of symmetry of the given parabola is [tex]2x-1=0.[/tex]
Step-by-step explanation: We are given to find the line of symmetry for the parabola with the following equation :
[tex]y=-x^2+x+3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We know that
the STANDARD equation of a parabola is given by
[tex]y=a(x-h)^2+k,[/tex]
where the line of symmetry is x - h = 0.
From equation (i), we get
[tex]y=-x^2+x+3\\\\\Rightarrow y=-(x^2-x)+3\\\\\Rightarrow y=-\left(x^2-2\times x\times\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+3+\left(\dfrac{1}{2}\right)^2\\\\\\\Rightarrow y=-\left(x-\dfrac{1}{2}\right)^2+3+\dfrac{1}{4}\\\\\\\Rightarrow y=-\left(x-\dfrac{1}{2}\right)+\dfrac{13}{4}.[/tex]
Comparing with the standard form of the parabola, the line of symmetry is given by
[tex]x-\dfrac{1}{2}=0\\\\\Rightarrow 2x-1=0[/tex]
Thus, the required line of symmetry of the given parabola is [tex]2x-1=0.[/tex]
