Question

I need to know the vertex,intercepts, and range for this problem g(x)=2x^2-8x+6

Answer 1

First write g(x)=2x²-8x+6 as 2(x-2)²-2
g(x)=2(x-2)²-2
We can say that (x-2)² is always positive and y must has a min. value in vertex.
y=2(x-2)²-2
(x-2)² min value is 0 so x=2
y=-2
Thus its vertex is (2,-2)
For its range
(x-2)² always positive so (x-2)²=0 and x=2 and y=-2 so we can say that it goes ∞ and it end -2.
It's range (-2,∞).Because its vertex point is(2,-2)
Intercepts:
For y intercepts ,x=0
y=6
For x intercepts,y=0
0=2(x-2)²-2
1=(x-2)²
x-2=1 or x-2=-1
x=3 or x=1

Answer 2

`y=a(x-h)^2+k\\ \text{vertex}=(h,k)`


`g(x)=2x^2-8x+6\\ g(x)=2(x^2-4x+3)\\ g(x)=2(x^2-4x+4-1)\\ g(x)=2(x-2)^2-2\\\text{vertex}=(2,-2)`

The range of a quadratic function is:

`y\in(-\infty,k]\text{ for } a<0\\ y\in[k,\infty)\text{ for } a>0\\`
So the range of
`g(x)` is
`y\in[-2,\infty)`.

x-intercepts:

`2(x-2)^2-2=0\\ 2(x-2)^2=2\\ (x-2)^2=1\\ x-2=1 \vee x-2=-1\\ x=3 \vee x=1`

y-intercepts:

`y=2(0-2)^2-2\\ y=2\cdot4-2\\ y=6`