How do find the vertex and axis of symmetry for a quadratic equation #y=-x^2+2x-5#?

Question

Genesis Allen · Accepted Answer

vertex at #(1,-4)#
axis of symmetry: #x=1# By rewriting #y =-x^2+2x-5# into vertex form: #color(white)("XXXX")##y=m(x-a)^2+b# (which has a vertex at #(a,b)#) we can easily extract the vertex. #y=-x^2+2x-5# #color(white)("XXXX")#extract #m# #y = (-1)(x^2-2x) -5# #color(white)("XXXX")#complete the square #y=(-1)(x^2-2x+1) -5 +1# #color(white)("XXXX")#simplify into vertex form #y=(-1)(x-1)^2+ (-4)# with a vertex at #(1,-4)# Since this is a parabola in standard position (vertical axis) the axis of symmetry runs though the x coordinate of the vertex.

Abigail Allen · Answer

undefined To find the vertex of a quadratic equation in the form $ y = ax^2 + bx + c $, use the formula:
$$ x = -\frac{b}{2a} $$
Once you find the value of $ x $, substitute it back into the original equation to find the corresponding value of $ y $. This will give you the coordinates of the vertex. The axis of symmetry is the vertical line that passes through the vertex. It's the line represented by the equation $ x = \frac{-b}{2a} $.

For the given equation $ y = -x^2 + 2x - 5 $:
$$ a = -1, \quad b = 2 $$
$$ x = \frac{-2}{2*(-1)} = 1 $$
$$ y = -(1)^2 + 2*(1) - 5 = -1 + 2 - 5 = -4 $$
So, the vertex is at $ (1, -4) $ and the axis of symmetry is $ x = 1 $.