Question

What are the domain and range of the function y=2√3x+4-5?

Answer 1

Is this your question
\[\large y= 2\sqrt{3x+4}-5\]

Answer 2

yes

Answer 3

Square root can't have a negative number inside
so
\[3x+4\ge 0\]
or
\[x\ge -\frac{4}{3}\]
so Domain of y is \[[-\frac{4}{3}, \infty]\]

Answer 4

minimum value of y will occur when x=-4/3, that's when the root will be zero
\[y=0-5=-5\]
so this is the minimum value of y, as x increases, y will increase
so range of y is
\[\large [-5, \infty)\]

Answer 5

?

Answer 6

Sorry
the domain will have the round bracket to the right side
\[[-\frac{4}{3}, \infty)\]

Answer 7

it isn't clear where the square root is suppose to end, but 2*sqrt(3x+4)-5 seems more likely than 2*sqrt(3x+4-5) or 2*sqrt(3x)+4-5.

The only requirement we have for x is that it makes the number in the radical >= 0:

3x+4 >= 0
3x >= -4
x >= -4/3

The range is the set of numbers that y could be. The smaller x is, the smaller y is, so the smallest y can be is when x is as small as it can be:

x = -4/3 makes y = 2*0 - 5 = -5

y can be bigger if x gets bigger, so y >= -5

Answer 8

now do you get it?

Answer 9

yes

Answer 10

thank you

Answer 11

welcome :)