Question

Given the function h(x) = 4^x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

Answer 1

Ooh, fun

the average rate of change is the slope

easy, so for section A, it would be the slope from (0,h(0)) to (1,h(1)) or from (0,1) to (1,4)
slope=(y2-y1)/(x2-x1)=(4-1)/(1-0)=3/1=3


section B, would be slope from (2,h(2)) to (3,h(3)) or from
(2,16) to (3,64)
slope=(y2-y1)/(x2-x1)=(64-16)/(3-2)=48/1=48


1. section A has an average rate of change of 3
section B has an average rate of change of 48



2. 48 is how many times greater than 3
48/3=16
16 times greater

it is because it is an exponential function and the slope grows bigger

Answer 2

Heyo Ace Baby!

First of all, a rate of change can be expressed as a slope. A slope is the ratio of the vertical and horizontal changes between two points. Average rate is always equal for every point and the line is straight.

So, our function is h(x) = $4^x$

we substitute the values and remember, x = 0 is the rise and x =1 is the run, our changes applies to x and h(x)

h(x) = 4^0 = 1 and 4^1 = 4

So, with the formula ($y2-y1$)/$x2-x1$) you can find the average rate (and let's say that you can find it also with (changes of h(x)) / (changes of x)). h(x) is considered as "y"

Section A = $\frac{4-1}{1-0} = \frac{3}{1}$ = 3

Let's work out the Section B...4^2 = 16 and 4^3 = 64

Section B = $\frac{64-16}{3-2} = \frac{48}{1}$ = 48

Let's find how bigge ris 48 than 3

48 : 3 = 16

16 times, because our function is exponential and slope is bigger in section B