What is the axis of symmetry and vertex for the graph #y = -2x^2 + 10x - 1#?

Question

Eli Assouline · Accepted Answer

Axis of symmetry is #x-5/2=0# and vertex is #(5/2,23/2)#

To find axis of symmetry and vertex, weshould convert the equation to its vertex form #y=a(x-h)^2+k#, where #x-h=0# isaxis of symmetry and #(h,k)# is the vertex.

#y=-2x^2+10x-1#

#=-2(x^2-5x)-1#

#=-2(x^2-2xx5/2xx x+(5/2)^2)+2(5/2)^2-1#

#=-2(x-5/2)^2+23/2#

Hence axis of symmetry is #x-5/2=0# and vertex is #(5/2,23/2)#

graph{(y+2x^2-10x+1)(2x-5)((x-5/2)^2+(y-23/2)^2-0.04)=0 [-19.34, 20.66, -2.16, 17.84]}

Jace Anderson · Answer

undefined The axis of symmetry for the graph $ y = -2x^2 + 10x - 1 $ is $ x = \frac{-b}{2a} $, where $ a = -2 $ and $ b = 10 $. Therefore, $ x = \frac{-10}{2*(-2)} = \frac{-10}{-4} = \frac{5}{2} $. The vertex corresponds to the point $ \left(\frac{5}{2}, f\left(\frac{5}{2}\right)\right) $. Substituting $ x = \frac{5}{2} $ into the equation yields $ y = -2\left(\frac{5}{2}\right)^2 + 10\left(\frac{5}{2}\right) - 1 = -2\left(\frac{25}{4}\right) + \frac{50}{2} - 1 = -\frac{25}{2} + 25 - 1 = -\frac{25}{2} + 24 = -\frac{1}{2} $. Therefore, the vertex is $ \left(\frac{5}{2}, -\frac{1}{2}\right) $.