Question

Identify the axis of symmetry, vertex, and y intercept of y=-6x^2+12x-8

Answer 1

See below for answers and explanations (with attached graph)

Explanation:

It helps to transform the equation into vertex form by completing the square because it tells us a lot about the characteristics of the parabola:

`\displaystyle y=-6x^2+12x-8\\\\y=-6\biggr(x^2-2x+(4)/(3)\biggr)\\\\y-6\biggr(-(1)/(3)\biggr) =-6\biggr(x^2-2x+(4)/(3)-(1)/(3)\biggr)\\\\y+2=-6(x^2-2x+1)\\\\y+2=-6(x-1)^2\\\\y=-6(x-1)^2-2`

Since vertex form is
`y=a(x-h)^2+k`, we identify the vertex to be
`(h,k)\rightarrow(1,-2)` and the axis of symmetry to be
`x=h\rightarrow x=1`. The y-intercept can be found by setting
`x=0` and evaluating:

`y=-6(x-1)^2-2\\\\y=-6(0-1)^2-2\\\\y=-6(-1)^2-2\\\\y=-6(1)-2\\\\y=-6-2\\\\y=-8`

Hence, the y-intercept of the parabola is
`y=-8`, or
`(0,-8)` as an ordered pair.