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Revision:Curve Sketching

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Curve Sketching

When you are asked to sketch a curve, you need be able to draw a quick sketch of the curve, showing the main details (such as where the curve crosses the axes). You should be able to quickly sketch straight-line graphs, from your knowledge that in the equation y = mx + c, m is the gradient and c where the graph crosses the y-axis.

When asked to sketch a more complicated graph, there are a number of things that you should work out before drawing your sketch.

Asymptotes. these are lines for which the graph is undefined. Remember that you cannot divide by zero. Therefore, in the graph of y = 1/(1 + x), x = -1 is an asymptote because when x is -1, you end up dividing by zero. What happens as the curve approaches an asymptote? Which way does it approach from?
Where the graph crosses the axes. The graph will cross the x-axis when y = 0 and the y-axis when x = 0. Substitute in x = 0 and then y = 0 to determine the crossing points, and mark these on your graph.
What happens as x becomes very large? Think about whether y will become very large, very small, positive or negative. What happens as x becomes very large and negative?
Is the graph symmetrical about the x or y axes? Remember, the graph is symmetrical about the y-axis if replacing x by -x in the equation of thegraph doesn't change the equation. The graph is symmetrical about the x-axis if replacing x by -x does not change the equation of the graph, apart from making the equation the negative of the original equation.
You may also think about where the maxima and minima occur (by differentiating).

Example

Sketch $y = \dfrac{1+x}{1-x}.$

When x = 1, we end up dividing by zero so there will be an asymptote at x = 1.
We can also rearrange this equation to get:

$y-1 = x(y+1) \Rightarrow x = \dfrac{y-1}{y+1}.$
This shows that there is also an asymptote at y = -1.
When x = 0, y = 1. Therefore the curve crosses the y-axis at (0, 1). When y = 0, 1 + x = 0 so x = -1. Therefore the curve crosses the x-axis at (-1, 0).
$y = \dfrac{1+x}{1-x} = -1 + \dfrac{2}{1-x}$ (which can be easily verified using polynomial division). As x gets large and positive, the fraction will become small and negative (2 divided by a large, negative number), so y tends towards -1 (from the 'bottom'). As x gets large and negative, the fraction will become small and positive (2 divided by a large, positive number), so y tends towards -1 (from the 'top').
$x = 1 - \dfrac{2}{y+1}$ (which can again be easily verified). As y gets large and positive, x tends to 1 from the left; as y gets large and negative, x tends to 1 from the right.
By substituting in -x for x it can be seen that the graph is not symmetrical in the x or y axes.