Please help very urgent!! how do you find the growth rate (like the percentage of growth each week) for an exponential function when you only have the starting value, the amount of weeks and the ending value (for example, a plant starts at (0,2) when it is planted, after 18 weeks it is 20 inches) how do you find out the percent of growth each week to plug it into an equation???

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  the n-th root of the ratio of ending to starting values, less 1

Step-by-step explanation:

When I have starting and ending values over a time period, I usually use those directly in the exponential equation. If you're asked for a percentage growth rate, then you have to convert the growth factor to a growth rate.

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The exponential equation can be written ...

  f(t) = (initial value) × ((final value)/(initial value))^(t/(time period))

In your example case, this would be ...

  f(t) = 2 × (20/2)^(t/18)

  f(t) = 2(10^(t/18)) . . . . . . simplified

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To convert this to the form ...

  f(t) = 2(1 +(growth rate))^t

recognize that ...

  (1 +(growth rate)^t = (10^(1/18))^t

So, you need to find the value of ...

  10^(1/18) = 10^(1/18) ≈ 1.13646 = 1 + 13.646%

Then the growth rate is about 13.6% each week.

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Sometimes, you want to use this in the form ...

  f(t) = (initial value)×(1 + (growth rate))^t

and sometimes you want the form ...

  f(t) = (initial value)×e^(kt)

When you solve for k, you find it is ...

  k = ln(1 + (growth rate)) . . . . . . . . where ln(x) is the natural logarithm of x

Here, that is ...

  k = ln(10^(1/18)) = ln(10)/18 ≈ 0.12792

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Your final growth equation could be any of ...

  f(t) = 2(10^(t/18)) . . . . . . . my personal favorite, as no rounding is involved

  f(t) = 2(1.13646^t)

  f(t) = 2e^(0.12792t)

y = 5x + 17

step-by-step explanation:

a perpendicular line will be the opposite reciprocal slope of the given equation.   the slope given is -1/5 so the opposite reciprocal is 5

find b:

7 = 5 (-2) + b

7 = -10 + b

17 = b

write equation

y = 5x + 17