Method for sketching $y = (1 – 3t + 2t^{2})e^{3t}$

Question

I am doing some examination practice, and I've faced the following question:

Another particular solution which satisfies $y = 1$ and $\frac {dy}{dx} = 0$ when $t = 0$, has equation $$y = (1 – 3t + 2t^{2})e^{3t}$$ For this particular solution draw a sketch graph of y against t, showing where the graph crosses the t-axis.

However I am having trouble drawing the graph. I know how to draw the two parts individually but the only thing I can think of when drawing $y = (1 – 3t + 2t^{2})e^{3t}$ is that $(1 – 3t + 2t^{2})$ will grow exponentially bigger every time.

What method or thinking strategy can I use to draw that graph?

(answer)

Answer 1

Presumably the only things they care about in the exam are:

a) Where are the values positive, where negative?

b) What is $f(0)$?

c) Where are the zeros, maxima and minima, are there saddle points?

d) What is the behaviour for $x\to\pm\infty$

e) Is the function continuous or even smooth?

Most of these questions can be solved very elementary and some might even be trivial. As soon as you solved those questions you know how to draw it, it doesn't matter if you hit the value for an arbitrary point, say $\frac 78$ correctly.

Did I miss any important properties?

Edit: In your particular example you will find $f(0)=1$, you will find two zeros ($1$ and $\frac 12$), a maximum and a minimum (derive!) the graph is smooth and goes to infinity for $x\to \infty$ quite rapidly and to $0$ for $x\to - \infty$. So just draw these points and connect the dots smoothly.

Answer 2

A useful technique for producing a graph of this type without using software is this:

Draw the two graphs on the same axes and the work out what the product of the two graphs would look like using the following principles:

• at points where one graph is zero (crosses the x-axis), the product graph is zero.
• in intervals where both graphs are non-zero, if both graphs are above (or both below) the x-axis, then the product graph will be above the x-axis, otherwise below the x-axis.

This approach is especially useful for graphs like $y = \sin{x}e^x$, where the curve oscillates between touching $y=e^x$ and $y=e^{-x}$, but works OK for your example.

Answer 3

To draw graphs of "nice" functions like that, there are a couple of steps:

1. Find where the function hits the $t$-axis($y=0$) and y-axis ($t=0$) and where it is positive and negative.

2. Find, if any, the vertical(the $t$ values where $y$ is undefined) and horizontal ($\lim_{t \rightarrow \infty} y $) asymptotes.

3. Find where the maxima/ minima ($\frac{d y}{d t}=0$) are and where the function is increasing ($\frac{d y}{d t} >0$) and decreasing ($\frac{d y}{d t}<0$).

4. One can then find the points of inflexion (where $\frac{d^2 y}{dt^2} =0$), these are just where the function changes from being concave up to concave down and vice versa.

Then put it all together and try drawing it!