y=5^2x-1 +2, what is the domain, range, horizontal asymtop,behavior, whats h & k, and the X & Y intercepts?

Answer 1

y=5^2x-1 +2

Assuming you mean

y=5^(2x-1) +2

y - 2 = 5^(2x - 1)
= 5^(2(x - ½))
= 25^(x - ½)

(y - h) = 25^(y - k)
h = 2; k = ½

Y = 25^X
where X = x - ½ and Y = y - 2

For Y = 25^X
Domain: All X; Range Y>0

So for y=5^(2x-1) +2
Domain: Still all x; Range y - 2 > 0 ie y >2)

Now Y = 25^X is assymptotic to the Y axis
So y=5^(2x-1) +2 is assymptotic to y = 2 (where the Y-axis is under the (x.y) frame of reference)

Since its assymptotic to y = 2 and range is y>2 then it NEVER cuts the x-axis (y = 0)

Cuts the y axis when x = 0
ie y = 5^(0 - 1) + 2
= 2.2

Summary
h = 2
k= ½
x intercept does not exist
y intercept = 2.2
Domain: All x (or -∞ < x < ∞)
Range: y > 2
Curve assymptotic to y = 2

Answer 2

if you mean:
y=5^{2x-1} +2
then
domain:=all real numbers
range y>2
horizontal asymptote: y=2
i do not know what h or k are
x=intercept: y=0, but, since the range is y>2,
that means that there is NO x-intercept
y-intercept: x=0, y = 5^{-1}+2 = 1/5 +2 =11/5

Answer 3

I assume you mean 5^(2x-1) + 2.

Domain: all real numbers, -infinity < x < infinity.
Range: y > 2, because an exponential function with a positive base is always positive, so 5^(2x-1) > 0 for all x.

Horizontal asymptote: y=2, because lim (x->-infinity) 5^(2x-1) = 0.

y-intercept: plug in x=0, and you get 5^(-1) + 2 = 11/5.
x-intercept: none

h,k? this isn't a parabola or other conic section, so I'm not sure what you're referring to in this case.

other behavior: y -> infinity as x -> infinity.