Question

1. $y = 2(2^x - 1)$

Explanation:

Step1: Simplify the function

$y=2(2^x - 1) = 2^{x+1} - 2$

Step2: Find y-intercept (x=0)

Substitute $x=0$:
$y=2^{0+1}-2=2-2=0$

Step3: Find x-intercept (y=0)

Set $y=0$:
$0=2^{x+1}-2 \implies 2^{x+1}=2^1 \implies x+1=1 \implies x=0$

Step4: Calculate key points

For $x=1$: $y=2^{2}-2=4-2=2$
For $x=-1$: $y=2^{0}-2=1-2=-1$
For $x=2$: $y=2^{3}-2=8-2=6$
For $x=-2$: $y=2^{-1}-2=\frac{1}{2}-2=-\frac{3}{2}$

Step5: Identify asymptote

As $x \to -\infty$, $2^{x+1} \to 0$, so $y \to -2$ (horizontal asymptote $y=-2$)

Step6: Plot points and draw curve

Plot $(0,0)$, $(1,2)$, $(-1,-1)$, $(2,6)$, $(-2,-\frac{3}{2})$, then draw a smooth curve approaching $y=-2$ as $x$ decreases.

Answer:

The graph is an exponential curve passing through $(0,0)$, $(1,2)$, $(-1,-1)$, approaching the horizontal asymptote $y=-2$ as $x \to -\infty$, increasing for all real $x$.