-4x^2+8x-1?

Question

what is the vertex (using completing the square method),
line of symetry,
x intercepts,
y intercepts,
domain,
range

Answer 1

-4x^2+8x-1=y

Completing the square:
-4(x^2-2x+1) +4 -1=0
-4(x-1)^2 +3 =0 So x=1 at the vertex and
y=3 , (1,3)

Line of symmetry is the value of x at the vertex, x=1. Also
the line of symetry is where x=-b/(2a) , in this case is -8/((2)(-4))=
-8/-8=1 , x=1

x intercepts are the roots or solutions to the equation which
are 1(+-)√(3)/2 (I used the quadratic formula)

y intercept is where x=0 and that's at y=-1

the domain is what x is allowed to be and that's anything

the range is what y is allowed to be - y opens downward. You
can see that by allowing x to become very large positively or negatively. The result is that y gets large in a negative direction.
The highest y can be is on the line of symmetry which is where
x=1. Letting x=1 in the equation we have
y=-4(1^2) + 8(1) -1= -4+8 -1= 3. So y<=3(the range)

Answer 2

Presuming you meant y = -4x^2 + 8x - 1

To find the vertex, you complete the square

y = -4 (x^2 - 8x) - 1
y = -4 (x^2 - 8x + 16) - 1 + 48
y = -4 (x - 4)^2 - 47

Vertex is at (4, -47). You take the negative of the number in brackets for the x-coordinate., and the number added/subtracted outside of the brackets is the y-coordinate.

x-intercept: defined to be where y = 0. So make y = 0 and solve for x.
0 = -4(x-4)^2 - 47
47 = -4(x-4)^2
47/-4 = (x-4)^2

There are no x-intercepts because (x-4)^2 must be a positive number.
To find the y-intercepts, make x = 0, and the answer should be -1.

The domain is all real numbers.
The range is the vertex downward. In this case, the y-coordinate of the vertex is -47, so the range is (-infinity,-47]

Answer 3

vertex : x = -b/2a = 1 ; y = 3 V(1 ; 3)
line of symmetry x = 1
x intercepts : plug x=0 , we have y = -1 : A(0 ; -1)
y intercepts : solve the equation -4x^2+8x-1 = 0 we have
x = (-2 + square root of 3)/-2 or x = (-2 - square root of 3)/-2