Find the axis of symmetry and vertex for the parabola y=−5x^2+10x−6.

Revision
- General equation of a parabola:
[tex]y=a\cdot(x-h)^2+k[/tex]- Coordinates of the vertex of a parabola:
[tex](h,k)[/tex]- Axis of symmetry of the parabola:
[tex]x=h[/tex]Answer
We have the following parabola:
[tex]y=-5x^2+10x-6[/tex]We can read the coordinates of the vertex and axis of symmetry from the equation of a parabola if we express its equation in the general form above. To do that we will "complete squares" in the following way:
[tex]\begin{gathered} y=-5(x^2-2x)-6 \\ y=-5(x^2-2x+1-1)-6 \\ y=-5(x^2-2x+1)+5-6 \\ y=-5(x-1)^2-1 \end{gathered}[/tex]Comparing this equation with the general equation of the parabola, we see that:
[tex]\begin{gathered} h=1 \\ k=-1 \end{gathered}[/tex]So the coordinates of the vertex are:
[tex](1,-1)[/tex]and the axis symmetry is:
[tex]x=1[/tex]