Question Video: Matching Exponential Functions to Given Data | Nagwa Question Video: Matching Exponential Functions to Given Data | Nagwa

# Question Video: Matching Exponential Functions to Given Data Mathematics • Second Year of Secondary School

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The given table shows the value of the function π at various inputs. Which expression for π(π) best fits the data? [A] 230(1.12)^(1 β π) [B] 230(0.79)^(π) [C] 230(0.79)^(βπ) [D] 230(1.12)^(π) [E] 230(0.79)^(π β 1).

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### Video Transcript

The given table shows the value of the function π at various inputs. Which expression for π of π best fits the data? (A) 230 multiplied by 1.12 to the power of one minus π, (B) 230 multiplied by 0.79 to the power of π, (C) 230 multiplied by 0.79 to the power of negative π, (D) 230 multiplied by 1.12 to the power of π, or (E) 230 multiplied by 0.79 to the power of π minus one.

Now to help us solve this problem, what we do is remember our general form for an exponential equation. And that is that π of π₯ is equal to π΄ multiplied by π to the power of π₯, where π΄ is the initial amount, π tells us about the rate, and then π₯ is our independent variable, and that is usually time. And we also have some conditions, and that is that π must be positive and must not be equal to one. And we can see that all of our answers are in this form.

And in fact for our problem, what weβre gonna have is the form π of π equals π΄π to the power of π. A nd thatβs because π is our independent variable. Well, now, in order to solve this type of problem where weβve been given different data points, what we look out for is a data point in the form zero, π΄. And the reason we look out for this is because what this does is tells us our initial amount. And thatβs because it tells us the value of our function at zero. And in this case, itβll be π equals zero.

So therefore, for this example, we can say that π΄ is equal to 230. And now what weβre gonna do is use another data point. And what weβre gonna do is use this data point to help us find our other variables. So, weβre gonna use the second data point along, where π equals two and π of π equals 140. Well, then, if we substitute this into our general form, what weβre gonna have is 140. Itβs because thatβs our π of π is equal to 230, which is our π΄ which we calculated, multiplied by π. And then itβs squared. And thatβs because our π-value is two.

Well, then what we can do is divide both sides by 230. And when we do this, we get 140 over 230 equals π squared. Well, then, what we can do is take the square root of both sides of our equation to find π. And when we do that, we get 0.780 et cetera is equal to π. And then if we remember, we donβt need to worry about the negative result because weβve already said that π must be positive and not equal to one. Well, then, if we put this back into our function of π, weβre gonna get π of π is equal to 230 multiplied by and weβve got 0.780 etcetera to the power of π. Well, then, if we check this against our possible answers, we can see that the closest answer is (B) 230 multiplied by 0.79 to the power of π.

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