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

Nov 16, 2017

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$\sqrt{x}$
$\left(- \frac{2}{3}\right)$=$\sqrt{x}$
$x$=$\frac{4}{9}$ Point ($\frac{4}{9}$, 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]}