Question

t(x) = 6(1.5)^{x + 2} - 4; v(x) = 6(1.5)^{x + 4} + 2; s(x) = 6(1.5)^{x + 4} - 2; j(x) = 6(1.5)^{x - 2} - 4

Explanation:

Step1: Analyze \( t(x) = 6(1.5)^{x + 2}-4 \)

Find the horizontal asymptote (HA) and a key point. The HA of an exponential function \( a(b)^{x - h}+k \) is \( y = k \). For \( t(x) \), \( k=-4 \), so HA is \( y = -4 \). Let \( x=-2 \), then \( t(-2)=6(1.5)^{0}-4 = 6 - 4=2 \). So the graph should approach \( y=-4 \) and pass through \((-2,2)\).

Step2: Analyze \( v(x) = 6(1.5)^{x + 4}+2 \)

HA is \( y = 2 \) (since \( k = 2 \)). Let \( x=-4 \), \( v(-4)=6(1.5)^{0}+2=6 + 2 = 8 \). So the graph approaches \( y = 2 \) and passes through \((-4,8)\).

Step3: Analyze \( s(x) = 6(1.5)^{x + 4}-2 \)

HA is \( y=-2 \) ( \( k=-2 \) ). Let \( x=-4 \), \( s(-4)=6(1.5)^{0}-2=6 - 2 = 4 \). So the graph approaches \( y=-2 \) and passes through \((-4,4)\).

Step4: Analyze \( j(x) = 6(1.5)^{x - 2}-4 \)

HA is \( y=-4 \) ( \( k=-4 \) ). Let \( x = 2 \), \( j(2)=6(1.5)^{0}-4=6 - 4 = 2 \). So the graph approaches \( y=-4 \) and passes through \((2,2)\).

Now match with the graphs:

• Top - Left: Approaches \( y=-4 \), passes through \((-2,2)\) → \( t(x) \)
• Top - Right: Approaches \( y = 2 \), passes through \((-4,8)\) → \( v(x) \)
• Bottom - Left: Approaches \( y=-2 \), passes through \((-4,4)\) → \( s(x) \)
• Bottom - Right: Approaches \( y=-4 \), passes through \((2,2)\) → \( j(x) \)

Answer:

Top - Left: \( t(x) = 6(1.5)^{x + 2}-4 \)
Top - Right: \( v(x) = 6(1.5)^{x + 4}+2 \)
Bottom - Left: \( s(x) = 6(1.5)^{x + 4}-2 \)
Bottom - Right: \( j(x) = 6(1.5)^{x - 2}-4 \)