What is the range of-sqrt(2x-8)+1 ?

The range of-sqrt(2x-8)+1 is f(x)<= 1

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range of-sqrt(2x-8)+1

range

- 2 x - 8 + 1

f (x) \leq 1

Interval Notation

(- \infty, 1]

Solution steps

The range of an radical function of the form

- c a x + b + k

f (x) \leq k

k = 1

f (x) \leq 1

y = \frac{5}{7} x - 4

(2.4) 3

(- 2, - 4), (- 1, 5)

f (x) = \frac{2}{x} - \frac{x}{x + 2}

f (x) = 4 (x - 1)^{2} - 3