SOLUTION: Find the domain (with the two restrictions)of the function: y=(sqr(x+7))/(x-8) (i.e. square root of x + 7 on top the fraction bar with x - 8 on bottom) I got [-7, 8)U(8, infi

Algebra ->  Functions -> SOLUTION: Find the domain (with the two restrictions)of the function: y=(sqr(x+7))/(x-8) (i.e. square root of x + 7 on top the fraction bar with x - 8 on bottom) I got [-7, 8)U(8, infi      Log On


   

Question 809797: Find the domain (with the two restrictions)of the function:
y=(sqr(x+7))/(x-8)
(i.e. square root of x + 7 on top the fraction bar with x - 8 on bottom)
I got [-7, 8)U(8, infinity) but home work site says it's incorrect. Puzzled!
Found 2 solutions by josmiceli, josgarithmetic:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Here's a plot:
+y+=+sqrt%28+x%2B7+%29+%2F+%28+x-8+%29+
+graph%28+400%2C+400%2C+-20%2C+20%2C+-10%2C+10%2C+sqrt%28+x%2B7+%29+%2F+%28+x-8+%29+%29+
--------------
The domain looks like
+x+%3E=+-7+ because there can't be a square root
of a negative number in the numerator
But the point +x+=+8+ must be excluded since
the denominator can't be = +0+


Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
y=%28sqrt%28x%2B7%29%29%2F%28x-8%29

Numerator will not be less than zero, and denominator cannot be 0.

x%3E=-7 and x%3C%3E8.
If you could view that on a number line, you may be able to give the interval notation for the domain:
[-7,8)U(8,oo)
or just as you said, [-7,8)U(8,infinity)

You did give the correct answer for the function' domain.