Question

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

Answer 1

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}`