Aim: To sketch a non-standard graph by sketching each part separately then adding the ordinates together.

#### Example

… … Sketch $y=\dfrac{1}{(x+1)^2}+x-1$

Solution

By examining this equation, we see there will be

• a vertical asymptote with the equation $x = -1$ and
• an oblique asymptote with the equation $y = x - 1$.

We can divide the function into two sections
… … $y_1=\dfrac{1}{(x+1)^2}$ and

… … $y_2 = x-1$

Use a pencil or a light colour to sketch these two simple functions on the one set of axes.

• y1 is in blue, y2 is in green

Notice the vertical asymptote of the truncus at $x = -1$.

The domain of the truncus will therefore be $x \in R \backslash \{ -1 \}$

Now work across the graph, comparing and adding the y-coordinates of the two simple graphs for a variety of different x-coordinates. Plot each point as you find it (marked in red)

• at $x = -3, \qquad y_1 = 0.25, \quad y_2 = -4, \quad \text{ so } \quad y_1 + y_2 = -3.75$
• at $x = -2, \qquad y_1 = 1, \qquad y_2 = -3, \; \quad \text{ so } \quad y_1 + y_2 = -3$
• at $x = -1.5, \quad y_1 = 5, \qquad y_2 = -2.5, \quad \text{ so } \quad y_1 + y_2 = 2.25$
• etc

Note:

• The final value of $y_1 + y_2$ is always greater than the $y_2 = x - 1$ value because the truncus ($y_1$) is always positive.
• At each end of the graph, the truncus is only adding a tiny amount so the the final value of $y_1 + y_2$ is very close to $y_2$
• Therefore, the graph will be approaching (from above) the line $y_2 = x - 1$ as an oblique asymptote as $x \rightarrow +\infty \text{ and as } x \rightarrow -\infty$.
• The truncus has a vertical asymptote and is undefined at $x = -1$, so the sum of $y_1 + y_2$ is also undefined at that point.
• As x values approach the vertical asymptote, the values of the truncus become very large and dominate the sum of $y_1 + y_2$.
• Therefore the graph will approach $x = -1$ as a vertical asymptote, with $y \rightarrow +\infty \text{ for } x \rightarrow -1^+ \text{ and } x \rightarrow -1^-$
• The point $x = 1$ is important because it is the x-intercept for $y_2 = x - 1$.
• To the left of $x = 1, \; y_2 < 0$ so the final line is below the original truncus.
• To the right of $x = 1 \; y_2 > 0$ so the final line is above the original truncus.

Using all of this information, we can draw our final graph (in red).

… … Domain: $x \in R \backslash \{ -1 \}$

… … Range: $y \in R$

… … Asymptotes: $x = -1, \; y = x - 1$

… … y-intercept: $y = 0$

… … x-intercepts: $x = -1.62, \; 0, \; 0.62$

… … Local minimum at $(0.26, \; –0.11)$

Note:

• CAS calculator was used to find x-intercepts and stationary points.
• x-intercepts can be found by algebra (solving $y = 0$)
• stationary points can be found by calculus (solving $\dfrac {dy}{dx} = 0$)

## Summary

To draw a graph by addition of ordinates

1. Sketch (using pencil or a light colour) the two original graphs
2. Draw and label the asymptotes
3. For a variety of x-values, add the y-coordinates and mark points.
1. Usually you can estimate the y-values based on the sketch, rather than calculating each one
4. Pay particular attention to:
1. values close to any vertical asymptote(s)
2. each end of the x-axis
3. y-intercept
4. x-intercepts of either original graph
5. placement of x-intercepts and turning points
5. Sketch the resultant graph by drawing a smooth curve through the points.
6. Take care to draw the graph approaching the asymptotes (not touching or curling away)
7. Label graphs, axes, asymptotes, intercepts, turning points
1. you made need to use technology or calculus to find turning points and x-intercepts
8. State the domain, range and the equations of asymptotes

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