How do you write an exponential function whose graph passes through (0,-0.3) and (5,-9.6)?

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Answer 1 · 2016-11-03T17:06:39

Please see the explanation for steps leading to a solution.

An exponential function is:

$y = C e^{α (x)}$ $y = Ce^(alpha(x))$

We can find the value of C, using the point $(0, - 0.3)$ $(0, -0.3)$

$- 0.3 = C e^{α (0)}$ $-0.3 = Ce^(alpha(0))$

$e^{α (0)} = 1$ $e^(alpha(0)) = 1$ so we merely flip the equation and drop the exponential:

$C = - 0.3$ $C = -0.3$

We can use the point $(5, - 9.6)$ $(5, -9.6)$ to find the value of $α$ $alpha$ :

$- 9.6 = (- 0.3) e^{α (5)}$ $-9.6 = (-0.3)e^(alpha(5))$

Divide both side by $- 0.3$ $-0.3$

$\frac{- 9.6}{- 0.3} = e^{α (5)}$ $(-9.6)/(-0.3) = e^(alpha(5))$

$32 = e^{α (5)}$ $32 = e^(alpha(5))$

Take the natural logarithm of both sides, to make the exponential disappear:

$ln (32) = α (5)$ $ln(32) = alpha(5)$

Divide both sides by 5:

$α = \frac{ln (32)}{5}$ $alpha = ln(32)/5$

The above is the same as $ln (\sqrt[5]{32}) = ln (2)$ $ln(root(5)(32)) = ln(2)$

$α = ln (2)$ $alpha = ln(2)$

The exponential function is:

$y = (- 0.3) e^{ln (2) x}$ $y = (-0.3)e^(ln(2)x)$