Question

the linear function $g(x)$ passes through the points $(4, 2)$ and $(-2, 3.5)$. the exponential function $h(x)$ is shown on the graph.
which statement is true about the end behavior of functions $g(x)$ and $h(x)$?
options:
as $x$ approaches negative infinity, $g(x)$ approaches positive infinity and $h(x)$ approaches $-4$.
as $x$ approaches positive infinity, $g(x)$ approaches negative infinity and $h(x)$ approaches $-3$.
as $x$ approaches negative infinity, $g(x)$ approaches negative infinity and $h(x)$ approaches positive infinity.
as $x$ approaches positive infinity, $g(x)$ approaches negative infinity and $h(x)$ approaches negative infinity.

Explanation:

Step1: Find slope of \( g(x) \)

The slope \( m \) of a line through points \( (x_1,y_1)=(4,2) \) and \( (x_2,y_2)=(-2,3.5) \) is \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{3.5 - 2}{-2 - 4}=\frac{1.5}{-6}=-0.25 \). So \( g(x) \) has a negative slope.

Step2: Analyze end behavior of \( g(x) \)

For a linear function \( g(x)=mx + b \) with \( m<0 \), as \( x\to+\infty \), \( g(x)\to-\infty \); as \( x\to-\infty \), \( g(x)\to+\infty \).

Step3: Analyze end behavior of \( h(x) \)

From the graph, as \( x\to-\infty \), \( h(x)\to+\infty \); as \( x\to+\infty \), \( h(x)\to - 3 \) (horizontal asymptote? Or approaches -3).

Step4: Check options

• Option 1: As \( x\to-\infty \), \( g(x)\to+\infty \) (correct for \( g \)), \( h(x)\to - 4 \)? No, graph shows \( h(x)\to+\infty \) as \( x\to-\infty \). Wait, no, re - check. Wait the graph of \( h(x) \): when \( x\to-\infty \), the curve goes up (towards positive infinity), when \( x\to+\infty \), it approaches a horizontal line (maybe \( y = - 3 \) or so). Wait the first option says as \( x\to-\infty \), \( g(x)\to+\infty \) (correct, since slope negative) and \( h(x)\to - 4 \)? No. Wait the second option: as \( x\to+\infty \), \( g(x)\to-\infty \) (correct, slope negative) and \( h(x)\to - 3 \) (matches the graph's right - end behavior). Wait let's re - evaluate.

Wait the slope calculation: \( 3.5-2 = 1.5=\frac{3}{2} \), \( -2 - 4=-6 \), so \( m=\frac{3/2}{-6}=-\frac{3}{12}=-\frac{1}{4}=-0.25 \). So \( g(x) \) is decreasing. So as \( x\) increases (towards \( +\infty \)), \( g(x) \) decreases (towards \( -\infty \)). As \( x\) decreases (towards \( -\infty \)), \( g(x) \) increases (towards \( +\infty \)).

For \( h(x) \): from the graph, as \( x\to+\infty \), the function \( h(x) \) approaches a horizontal line (looks like \( y=-3 \) or so). As \( x\to-\infty \), \( h(x) \) goes to \( +\infty \).

Now check the options:

• Option 1: As \( x\to-\infty \), \( g(x)\to+\infty \) (correct), \( h(x)\to - 4 \) (incorrect, \( h(x)\to+\infty \) as \( x\to-\infty \)).
• Option 2: As \( x\to+\infty \), \( g(x)\to-\infty \) (correct, slope negative), \( h(x)\to - 3 \) (matches the right - end behavior of \( h(x) \) as it approaches a horizontal line around \( y = - 3 \)).
• Option 3: As \( x\to-\infty \), \( g(x)\to-\infty \) (incorrect, should be \( +\infty \) since slope negative), \( h(x)\to+\infty \) (correct for \( h \), but \( g \) is wrong).
• Option 4: As \( x\to+\infty \), \( g(x)\to-\infty \) (correct), \( h(x)\to-\infty \) (incorrect, \( h(x)\) approaches a constant, not \( -\infty \)).

Wait maybe I misread the graph. Wait the graph of \( h(x) \): when \( x\) is large positive, it's near \( y=-3 \). When \( x\) is large negative, it's going up (positive infinity). So the second option: "As \( x \) approaches positive infinity, \( g(x) \) approaches negative infinity and \( h(x) \) approaches \( - 3 \)." is correct.

Answer:

As \( x \) approaches positive infinity, \( g(x) \) approaches negative infinity and \( h(x) \) approaches \( - 3 \) (the second option in the list of options, which is: As \( x \) approaches positive infinity, \( g(x) \) approaches negative infinity and \( h(x) \) approaches \( - 3 \).)