How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of #y=-2x^2-8x+3#?

1 Answer
bp
Sep 29, 2015

Vertex(-2,11). Axis of symmetry x= -2
Y intercept is 3, x intercepts are #-2+sqrt22/2# and #-2-sqrt22/2#

Domain #{x in RR}#, Range #{yin RR, y<=11}#

Explanation:

Complete the square for x as follows:

#y= -2(x^2 +4x)+3#

#y= -2(x^2 +4x +4-4) +3#

#y= -2(x^2 +4x +4)+8 +3#

#y= -2(x+2)^2 +11#

This gives vertex (-2, 11) axis of symmetry x=-2,
For y intercept put x=0 in the given equation to get y=3. For x intercept put y= 0 to get #-2x^2 -8x+3=0#. Solve for x, using

quadratic formula x= #(8+-sqrt(64+24))/-4#

#-2-sqrt22/2, -2+sqrt22/2#.

The parabola opens downwards because of negative coefficient of #x^2#, has vertex at (-2,11), hence its domain would all of #RR# and range would be {y #in RR, y<=11}#

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