Question

How do you graph `y=sqrtx+1.5`, compare it to the parent graph and what is the domain and range?

Answer 1

Find some values for `x` that are easy to solve.
Domain: `[0,\infty)`
Range: `[1.5,\infty)`

Explanation:

Since `x>=0`, we can choose some points that are perfect squares.
Let's choose `x=0,1,4,9,16`
For these values we get the following
`f(0)=\sqrt{0}+1.5=0+1.5=1.5`
`f(1)=\sqrt{1}+1.5=1+1.5=2.5`
`f(4)=\sqrt{4}+1.5=2+1.5=3.5`
`f(9)=\sqrt{9}+1.5=3+1.5=4.5`
`f(16)=\sqrt{16}+1.5=4+1.5=5.5`

We can then use the following points to graph the line: `(0, 1.5),(1,2.5),(4,3.5),(9,4.5),(16,5.5)`
graph{sqrt(x)+1.5 [-1.71, 18.29, -1.2, 8.8]}

Compared to the parent graph, this function is moved up 1.5 units because the value is added after `x` has been "modified".

To find the domain we have to make sure there is no case for divide by zero and no negatives under radicals. Since nothing is divided, we don't have to worry about divide by zero. However, we do have a radical so we have to make sure whatever is in the radical is not negative.
`x>=0`
There we go! The domain is simply `[0,\infty)`.

To get the range we can just plug these values into the function and get `[1.5,\infty)`.