Question

graph the function.
$f(x)=2\sqrt{x + 3}$
plot four points on the graph of the function: the leftmost point and three additional points. then click on the graph - a - function button.

Explanation:

Step1: Find leftmost point (domain start)

The expression under the square root must be non-negative: $x+3\geq0 \implies x\geq-3$.
At $x=-3$: $f(-3)=2\sqrt{-3+3}=2\sqrt{0}=0$.
Leftmost point: $(-3, 0)$

Step2: Calculate point at $x=-2$

Substitute $x=-2$ into the function:
$f(-2)=2\sqrt{-2+3}=2\sqrt{1}=2(1)=2$.
Point: $(-2, 2)$

Step3: Calculate point at $x=1$

Substitute $x=1$ into the function:
$f(1)=2\sqrt{1+3}=2\sqrt{4}=2(2)=4$.
Point: $(1, 4)$

Step4: Calculate point at $x=6$

Substitute $x=6$ into the function:
$f(6)=2\sqrt{6+3}=2\sqrt{9}=2(3)=6$.
Point: $(6, 6)$

Answer:

Leftmost point: $(-3, 0)$
Additional point 1: $(-2, 2)$
Additional point 2: $(1, 4)$
Additional point 3: $(6, 6)$
(Plot these points and draw a smooth curve starting at $(-3,0)$ and increasing through the other points to graph $f(x)=2\sqrt{x+3}$)