For an exponential growth equation of the form y=a(b)x, a is the yy-intercept and bb is the growth factor.

The initial value is typically when x=0. If the initial value you are given is when x=0, then you can plug in its value for aa in the equation. This is because the y-intercept is when x=0. That is, plugging in x=0 into y=a(b)x gives y=ab0=a(1)=a.

For example, if you are told a city has an initial population of 50,000, then you would plug in a=50,000 since 50,000=a(b)0 gives a=50,000.

If the initial value given is not when x=0, then you need to plug its value in for yy and its corresponding value for xx into the equation.

For example, if you are given a city's population of 50,000 in 2001 but xx represents the number of years since 2000, then you would need to substitute y=50,000 and x=1 into the equation. This would give you the equation 50,000=ab. To find aa, you would need to then write a second equation using a second given point or using the growth factor to find the value of bb if it is given.

If you were given a growth rate of 2%, then b=100%+2%=102%=1.02. Substituting b=1.02 into ab=50,000 gives a(1.02)=50,000. You can then divide both sides by 1.02 to get a≈49,020.

If you were not given a growth rate but were given a second point, such as a population of 60,000 in the year 2002, then you could plug in x=2 and y=60,000 into y=a(b)x to write a second equation. This would give you the equation 60,000=ab2. You would then have the

system \(\displaystyle{\left\lbrace{a}{b}={50000},{a}{b}^{{2}}={60000}\right\rbrace}\) that you could then solve for aa and bb using the substitution method.

The initial value is typically when x=0. If the initial value you are given is when x=0, then you can plug in its value for aa in the equation. This is because the y-intercept is when x=0. That is, plugging in x=0 into y=a(b)x gives y=ab0=a(1)=a.

For example, if you are told a city has an initial population of 50,000, then you would plug in a=50,000 since 50,000=a(b)0 gives a=50,000.

If the initial value given is not when x=0, then you need to plug its value in for yy and its corresponding value for xx into the equation.

For example, if you are given a city's population of 50,000 in 2001 but xx represents the number of years since 2000, then you would need to substitute y=50,000 and x=1 into the equation. This would give you the equation 50,000=ab. To find aa, you would need to then write a second equation using a second given point or using the growth factor to find the value of bb if it is given.

If you were given a growth rate of 2%, then b=100%+2%=102%=1.02. Substituting b=1.02 into ab=50,000 gives a(1.02)=50,000. You can then divide both sides by 1.02 to get a≈49,020.

If you were not given a growth rate but were given a second point, such as a population of 60,000 in the year 2002, then you could plug in x=2 and y=60,000 into y=a(b)x to write a second equation. This would give you the equation 60,000=ab2. You would then have the

system \(\displaystyle{\left\lbrace{a}{b}={50000},{a}{b}^{{2}}={60000}\right\rbrace}\) that you could then solve for aa and bb using the substitution method.