Question

Consider the quadratic function shown in the table below. x y 0 0 1 3 2 12 3 27 Which exponential function grows at a faster rate than the quadratic function for 0< x < 3?

Answer 1

So values are

• (0,0)
• (1,3)
• (2,12)
• (3,27)

The rule is

• 3/1=3
• 12/2=6
• 27/3=9

Explicit formula

• x(3x) where x is set of integers

Equation here is

• y=3x²

Answer 2

Answer:

$f(x)=4^x$ grows at a faster rate than the given quadratic function.

Step-by-step explanation:

Given table:

$\large \begin{array}{| c | c |}\cline{1-2} x & y \\\cline{1-2} 0 & 0 \\\cline{1-2} 1 & 3 \\\cline{1-2} 2 & 12 \\\cline{1-2} 3 & 27 \\\cline{1-2}\end{array}$

First Differences in y-values:

$0 \overset{+3}{\longrightarrow} 3 \overset{+9}{\longrightarrow} 12 \overset{+15}{\longrightarrow} 27$

Second Differences in y-values:

$3 \overset{+6}{\longrightarrow} 9 \overset{+6}{\longrightarrow} 15$

As the second differences are the same, the function is quadratic.

The coefficient of $x^2$ is always half of the second difference.

Therefore, the quadratic function is:

$f(x)=3x^2$

The average rate of change of function f(x) over the interval a ≤ x ≤ b is given by:

$\dfrac{f(b)-f(a)}{b-a}$

Therefore, the average rate of change for $f(x)=3x^2$ over the interval
0 ≤ x ≤ 3 is:

$\textsf{Average rate of change}=\dfrac{f(3)-f(0)}{3-0}=\dfrac{27-0}{3-0}=9$

An exponential function that grows at a faster rate than $f(x)=3x^2$ over the interval 0 ≤ x ≤ 3  is $f(x)=4^x$

$\large \begin{array}{| c | c | c | c | c |}\cline{1-5} x & 0 & 1 & 2 & 3 \\\cline{1-5} f(x)=4^x & 1 & 4 & 16 & 64\\\cline{1-5} \end{array}$

$\textsf{Average rate of change}=\dfrac{f(3)-f(0)}{3-0}=\dfrac{64-1}{3-0}=21$

As 21 > 9, $f(x)=4^x$ grows at a faster rate than $f(x)=3x^2$ over the interval 0 ≤ x ≤ 3.