Verified Answer :

konrad509

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

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

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

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

Related Questions