What is the axis of symmetry and vertex for the graph #y=2x^2-8x-10#?

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Answer 1 · 2017-08-12T12:06:30

Axis of symmetry is #x-2=0# and vertex is #(2,-18)#.

For #y=a(x-h)^2+k#, while axis of symmetry is #x-h=0#, vertex is #(h,k)#.

Now we can write #y=2x^2-8x-10# as

#y=2(x^4-4x+4)-8-10#

or #y=2(x-2)^2-18#

Hence, axis of symmetry is #x-2=0# and vertex is #(2,-18)#.

graph{(y-2x^2+8x+10)(x-2)=0 [-10, 10, -20, 20]}

Answer 2 · 2017-08-12T12:10:30

Vertex is at # (2,-18) # and axis of symmetry is # x =2#

# y = 2x^2 -8x -10 or y = 2(x^2-4x) -10 # or

#y = 2(x^2-4x+4) -8 -10 or y=2(x-2)^2 -18#

Comparing with standard form of equation

#y=a(x-h)^2+k ; (h,k) # being vertex we find here

# h=2 , k=-18 # So vertex is at # (2,-18) # .

Axis of symmetry is #x=h or x =2#

graph{2x^2-8x-10 [-40, 40, -20, 20]} [Ans]