joshtheyellowmamba
Answered

Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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 
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.