joshtheyellowmamba
Answered

Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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

Sagot :

konrad509
[tex]y=a(x-h)^2+k\\ \text{vertex}=(h,k)[/tex]

[tex]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)[/tex]

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

x-intercepts:
[tex]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 [/tex]

y-intercepts:
[tex]y=2(0-2)^2-2\\ y=2\cdot4-2\\ y=6[/tex]
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 
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.