How do you write an exponential function whose graph passes through the points (0,3) and (-1,6)?

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Answer 1 · 2016-11-17T18:27:27

Please see the explanation.

Given: $(0, 3) and (- 1, 6)$ $(0, 3) and (-1, 6)$

Using the equation:

$y = a e^{α (x)}$ $y = ae^(alpha(x))$

Substitute 0 for x and 3 or y:

$3 = a e^{α (0)}$ $3 = ae^(alpha(0))$

$a = 3$ $a = 3$

Substitute 3 for a:

$y = 3 e^{α (x)}$ $y = 3e^(alpha(x))$

Substitute -1 for x and 6 for y:

$6 = 3 e^{α (- 1)}$ $6 = 3e^(alpha(-1))$

Divide by 3

$2 = e^{α (- 1)}$ $2 = e^(alpha(-1))$

Natural log of both sides:

$ln (2) = - α$ $ln(2) = -alpha$

$α = ln (\frac{1}{2})$ $alpha = ln(1/2)$

Substitute $ln (\frac{1}{2})$ $ln(1/2)$ for $α$ $alpha$ :

$y = 3 e^{ln (\frac{1}{2}) (x)}$ $y = 3e^(ln(1/2)(x))$

Here is the graph:

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