Question

draw the graph of $f(x)=3^{x + 3}$

Explanation:

Step1: Rewrite function using exponent rule

$f(x) = 3^{x+3} = 3^x \times 3^3 = 27 \times 3^x$

Step2: Find key points (x,f(x))

For $x=-3$: $f(-3)=3^{-3+3}=3^0=1$
For $x=-2$: $f(-2)=3^{-2+3}=3^1=3$
For $x=-1$: $f(-1)=3^{-1+3}=3^2=9$
For $x=0$: $f(0)=3^{0+3}=3^3=27$ (off the given grid)
For $x=-4$: $f(-4)=3^{-4+3}=3^{-1}=\frac{1}{3}\approx0.33$

Step3: Identify asymptote

As $x\to-\infty$, $3^{x+3}\to0$, so horizontal asymptote: $y=0$

Step4: Plot points and asymptote

Plot $(-4, \frac{1}{3})$, $(-3,1)$, $(-2,3)$, $(-1,9)$; draw a curve rising to the right, approaching $y=0$ as $x\to-\infty$.

Answer:

The graph is an exponential curve with a horizontal asymptote at $y=0$, passing through points $(-4, \frac{1}{3})$, $(-3,1)$, $(-2,3)$, $(-1,9)$, and increasing rapidly to the right (with $f(0)=27$ lying outside the provided grid bounds).