Question

How do you graph `f(x) = 3sqrtx + 2`?

Answer 1

You take a few critical points, and graph.

Explanation:

Normally we choose the Y-intercept (when x=0) and the X-intercept (when y=0), and x=1, x=-1, x=2, etc. Close values to zero.

Let's calculate the y-intercept first.

x=0
3`sqrt(0)`+2=0+2=2 Point (0, 2)

y-intercept

y=0

0=3`sqrt(x)`+2
-2=3`sqrt(x)`
`(-2/3)`=`sqrt(x)`
`x`=`4/9` Point (`4/9`, 0)

x=1

3`sqrt(1)`+2=5 Point (1, 5)

x=-1

3`sqrt(-1)`+2= ? Point does not exist because square root
of a negative number does not exist.

Pretty soon we can see that the graph will start AFTER zero and nothing before. Then it will be a curve low to high as it progresses right. Graph below:

graph{3sqrt(x)+2 [-10, 10, -5, 5]}