# Addition of Ordinates

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.

- y
_{1}is in blue, y_{2}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

- Sketch (using pencil or a light colour) the two original graphs
- Draw and label the asymptotes
- For a variety of x-values, add the y-coordinates and mark points.
- Usually you can estimate the y-values based on the sketch, rather than calculating each one

- Pay particular attention to:
- values close to any vertical asymptote(s)
- each end of the x-axis
- y-intercept
- x-intercepts of either original graph
- placement of x-intercepts and turning points

- Sketch the resultant graph by drawing a smooth curve through the points.
- Take care to draw the graph approaching the asymptotes (not touching or curling away)
- Label graphs, axes, asymptotes, intercepts, turning points
- you made need to use technology or calculus to find turning points and x-intercepts

- State the domain, range and the equations of asymptotes

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