This App will help you to avoid any unwanted slip-ups in the exam. Although most of the reminders are common sense, but from the evidence students still need reminding of them. Read through the tips and take note of the most relevant ones before tackling your exam.
The curves of y = kxn (where n is a positive integer)
You must be able to recognize the various graphs of y = kxn, so here they are for you to learn...
A cubic curve, (where x3 is the highest power of x), has one of the following shapes:
For a positive x3:
For a negative x3:
The diagrams show the minimum and maximum values of y in that region of the curve.
These points can be calculated, as they are the points where the differential (remembering from above that the differential equals the gradient of the curve) of the equation of the curve equals zero. (See the differentiation topic for more information.)
The other point you should be able to identify is the point of inflection. At this point the differential of the equation (and thus the gradient) of the curve does equal zero, but in the case of a point of inflection the gradient does not change from positive to negative, but carries on as a positive or negative gradient.
A cubic curve can have various positions in relation to the x-axis. Therefore the number of times it crosses the x-axis can be:
But, there is always at least one point of intersection with the x-axis and so a cubic equation has at least one real root.
There are three features that are of interest when sketching a quadratic graph
The graph crosses the y-axis when x = 0. For instance, at the value of the constant in the equation.
The graph crosses the x-axis when y = 0. For instance, solve the quadratic = 0. The turning point is found by either completing the square or using differentiation.
Sketch the graph of y = x2 - 4x - 5.
Firstly, it crosses the y-axis when x = 0, and y = -5.
Secondly, it crosses the x-axis when y = 0. (For instance, crosses the y-axis at (0, -5))
- x2 - 4x - 5 = 0
- (x - 5)(x + 1) = 0
- x = 5 or x = -1
(For instance, crosses the x-axis at (5, 0), and (-1, 0))
Thirdly, we complete the square to get:
- y = x2 - 4x - 5
- y = (x - 2)2 - 4 - 5 or
- y = (x - 2)2 - 9
As the smallest (x - 2)2 can be is zero (when x = 2), the minimum value for y is -9 when x = 2. (For instance, it turns at (2, -9))
If we don't already know what a graph will look like we need to find its main features. These are:
- Where the graph crosses the y-axis, which is when x = 0. (i.e. at the constant).
- Where the graph crosses the x-axis. To find the roots (where the graph crosses the x-axis), we solve the equation y = 0.
- Where the stationary points are. The stationary points occur when the gradient is 0 (i.e. differentiate.) Whether there are any discontinuities.
- Are there any discontinuities? A discontinuity occurs when the graph is undefined for a certain value of x. This occurs when x appears in the denominator of a fraction (you can't divide by zero).
- What happens as x approaches ± ∞? When x becomes a large positive or a large negative number the graph will tend towards a certain value or pattern.
Now put all this information onto the graph and join up the points.
Sketch the graph:
If x = -3 then the denominator is zero. As we cannot divide by zero the graph is undefined, and there is a discontinuity at x = 3.
As x approaches + ∞, y approaches 2 (the -1 and +3 become insignificant). As x approaches - ∞, y approaches 2 as well. This means there is a horizontal asymptote (value that the graph tends towards) at y = 2.
So the final graph looks like this:
The following equation is a Cartesian equation, for a circle:
(x - a)2 + (y - b)2 = r2
where (a, b) is the centre of the circle and R is the radius.
You may have noticed the similarity between the equation for a circle and Pythagoras' theorem. This is because the equation for a circle is derived from Pythagoras' theorem.
Check this by copying the picture and:
- Drawing a right-angled triangle inside the circle with the line joining the centre to the general point (x, y) as the hypotenuse.
- Work out the lengths of the other two sides.
- Use Pythagoras' Theorem to make the formula.
If you got that to work you have just derived the general formula for a circle!
When you see the equation of a circle in formula sheets it may look like this:
x2 + y2 + 2fx + 2gy + c = 0.
This is a very similar equation to the one above, but re-arranged so that:
The circle's centre is at (-f, -g) and,
The radius equal to
(Check to see if you agree - you will need to complete the square!)
If a straight line touches a circle, and at the point it touches the circle it is at right angles to the radius of the circle, then that line is said to be at a tangent to the circle:
To find the equation of this line we need to know 2 pieces of information.
1. The point where the tangent meets the curve. We do this by solving the simultaneous equations for the line and the circle.
2. The gradient of the tangent. We do this by finding the gradient of the radius
and then we use the idea that the product of the gradients of perpendicular lines is -1.
For instance, the gradient of the tangent is
Once we know these we can use the formula: y - y1 = m (x - x1) to get the gradient of the tangent.
Note: For our diagram, the gradient of the line at a tangent to a circle =
If you require the equation of a tangent to a curve, then you have to differentiate to find the gradient at that point, and then use the formula, (y - y1) = m(x - x1), as before.
If you require the equation of the normal to the tangent of a curve, (i.e. the line perpendicular to the tangent), then follow the same procedure as above, remembering to use the fact that the gradient of the normal =
Find the equation of the normal to the curve y = 3x2 - 2x + 1 at the point (1,2).
Gradient = 6x - 2,
- so when x = 1, gradient = 4.
- so, gradient of normal = -¼
- so, equation of the normal is, y - 2 = -¼(x - 1)
- so, y = 2¼ - ¼x or 4y + x = 9
As with the intersection of two lines, we...
- Use substitution to solve the simultaneous equations.
- Rearrange them to form a quadratic equation.
- Solve the quadratic by factorising, or by using the quadratic formula.
- Find the y-coordinates by substituting these values into the original equations.
Note: This will give 2, 1, or 0 solutions for the x coordinate at the point(s) of intersection.
Find where the line y = 3x - 2 meets the curve y = x2 + x - 5.
Therefore the line and curve intersect at (3,7) and (-1,-5).
In summary: the points of intersection of two curves are found by solving the equations of the two curves simultaneously.
If the equation can be solved (i.e. it has real roots), then the curves meet.
If the equation can not be solved (there are no real roots), and the curves do not meet.