Please, for clarity, use " ^ " to denote exponentiation:
Correct format: x^4*y*(4) = y*x^2*(13)
This is an educated guess regarding what you meant to share. Please err on the side of using more parentheses ( ) to show which math operations are to be done first.
Your (x+y)2, better written as (x+y)^2, equals x^2 + 2xy + y^2, when expanded.
The question here is whether you can find this x^2 + 2xy + y^2 in your
"X4y(4) = yx2(13)"
Please lend a hand here. If at all possible obtain an image of the original version of this problem and share it. That's the only way to ensure that your helpers won't have to guess what the problem actually looks like.
4x is the answer for this one
Answer:
Vertex: (1,-1)
Symmetry: x = 1
Step-by-step explanation:
y = 4x² - 8x + 3
y = 4(x² - 2x) + 3
y = 4[x² - 2(x)(1) + 1² - 1²] + 3
y = 4(x - 1)² - 1
In y = a(x - h)² + k,
(h,k) is the vertex
And a vertical line passing through the vertex is the line of symmetry, i.e x = h.
Vertex: (1,-1)
Symmetry: x = 1
Remember the exponential law that says
therefor
the answer is C
