Question

a graphing calculator is recommended. graph the given functions on a common screen. explain how these graphs are related. (y = 2^{x},y = e^{x},y = 5^{x},y = 20^{x}) all of these graphs approach as (x
ightarrow-infty), all of them pass through the point ((x,y)=) ( ), and all of them are increasing and approach (infty) as (x
ightarrowinfty). the larger the base, the ---select--- the function increases for (x > 0), and the ---select--- it approaches 0 as (x
ightarrow-infty). 4. -/0.5 points starting with the graph of (y = e^{x}), write an equation of the graph that results from the following changes. (a) shifting 7 units downward (y=) (b) shifting 9 units to the right

Explanation:

Step1: Analyze limit as x approaches negative infinity

For exponential functions of the form $y = a^x$ where $a>1$, as $x\to-\infty$, $\lim_{x\to-\infty}a^x = 0$. So all of $y = 2^x,y=e^x,y = 5^x,y=20^x$ approach $0$ as $x\to-\infty$.

Step2: Find the y - intercept

For any exponential function $y=a^x$, when $x = 0$, $y=a^0=1$. So all these functions pass through the point $(x,y)=(0,1)$.

Step3: Analyze growth rate

For exponential functions $y=a^x$ with $a > 1$, the larger the base $a$, the faster the function increases for $x>0$, and the faster it approaches $0$ as $x\to-\infty$.

Step4: For part (a) - vertical shift

The rule for shifting a function $y = f(x)$ down by $k$ units is $y=f(x)-k$. Starting with $y = e^x$, shifting 7 units downward gives $y=e^x - 7$.

Step5: For part (b) - horizontal shift

The rule for shifting a function $y = f(x)$ to the right by $h$ units is $y=f(x - h)$. Starting with $y = e^x$, shifting 9 units to the right gives $y=e^{x - 9}$.

Answer:

For the first part:

• First blank: $0$
• Second blank: $(0,1)$
• Third blank: faster
• Fourth blank: faster

For part (a): $y=e^x - 7$
For part (b): $y=e^{x - 9}$