50 milligrams of a volatile substance is placed in a laboratory. Its mass, $M$ milligrams, $t$ minutes later, is given by $M = 120 - k e^{-\tfrac{1}{2}t}$, where $k$ is a constant.

1. Substitute the initial conditions into the model to show the value of $k$

At $t = 0$, $M = 50$.
$50 = 120 - k e^{-\frac{1}{2}(0)}$
$50 = 120 - k(1)$
$k = 120 - 50$
$k = 70$

1. Determine the key features of the curve

When $t = 0$: $M = 120 - 70e^{0} = 120 - 70 = 50$, so the curve starts at $(0,\,50)$.

As $t \to \infty$: $e^{-\tfrac{1}{2}t} \to 0$, so $M \to 120$. Hence $M = 120$ is a horizontal asymptote, and $M$ increases towards it.

2. Sketch the curve with correct shape and labelled features

Draw an increasing, concave-down curve for $t \geqslant 0$:

• starting at the $M$-intercept $(0,\,50)$,
• rising steeply at first then flattening,
• approaching the horizontal asymptote $M = 120$ (shown as a dashed line) without crossing it.

Label the intercept $50$ and the asymptote $M = 120$.