Question

Hi can u have me by sketching a graph for this? u have to find the domain and range i believe

Answer 1

The given expression is

`y=1+\sqrt[]{x+3}`

To sketch the graph, first, we have to evaluate the function when x = 0, 1, 2, 3, 4.

`\begin{gathered} y=1+\sqrt[]{0+3}=1+\sqrt[]{3}\approx2.7 \\ y=1+\sqrt[]{1+3}=1+\sqrt[]{4}=1+2=3 \\ y=1+\sqrt[]{2+3}=1+\sqrt[]{5}\approx3.2 \\ y=1+\sqrt[]{3+3}=1+\sqrt[]{6}\approx3.4 \\ y=1+\sqrt[]{4+3}=1+\sqrt[]{7}\approx3.6 \end{gathered}`

Let's form points with these values: (0, 2.7), (1, 3), (2, 3.2), (3, 3.4), and (4, 3.6). Now we, plot these points and draw the curve across them to have the graph of the expression.

As you can observe, the domain would be all real numbers greater or equal than x = -3.

`D\colon\lbrack-3,\infty)`

And the range would be all real numbers greater than or equal to 1.

`R\colon\lbrack1,\infty)`