f(x) = root(x - 4), find x and y-intercepts and D and R.....?

Answer 1

I wish that Raj, the supposedly top answerer, would think a little more before jumping in. Several of his recent answers have been partially or totally incorrect. I've only been looking for a few weeks so how far back do the mistakes go?

The x intercept is clearly x = 4. He correctly states that the domain is x >= 4 so how can the x intercepts be 2 or -2?

I assume that D and R means domain and range. The domain as already stated is x >= 4. The range is NOT all reals , it is y >= 0. This is because a function cannot have two y values for each x value. Therefore the square root is defined to be positive.

Answer 2

f(x) = root(x - 4),
► setting x=0; f(x) = root(0 - 4)= root( - 4)= +/-2i which is on imaginary axis and it's not real; we have no real interception by y axis;

► setting y=0; 0 = root(x - 4), x=4

►Domain => x-4 >= 0 ; x >= 4 because under the square root must be positive;

►Range => because the answer from a square root is always positive so the range
y>=0 or R+

Answer 3

to discover y intercept, merely plug in 0 for x and remedy for y: y = -2(0) + 4 = 4 y-intercept = 4 to discover the x intercept, plug in 0 for y and remedy for x: 0 = -2x + 4 2x = 4 x = 2 x-intercept = 2

Answer 4

y intercept:
x=0
y=√(0-4)
the graph does not intersect the y-axis Ans
x-intercept:
y=0
0=√(x-4)
x-4=0
x=4 Ans

Domain:
x-4≥0 (as the value under the √ shud be +ve)
x≥4
Range :
y≥0
as the roots are only +ve.

Answer 5

f(x)=rt(x-4)
setting x=0 y intercept=imaginary
setting y=0 x intercept=+/-2
what is D and R I don't know
can you clarify?
is it domain and range
domain x>=4
range all reals