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By solving an inequality on a graph that involves both x and y we get a region on a graph rather than a line. The region can be of any shape and does not need to be in any part of the graph. We can find the region by carrying out two steps:
1) Change the inequality to an equation (by changing ‘more than’ and ‘less than’ signs to equals) and then plot the line of this equation on a graph. This is called the boundary line.
2) Decide which of this line will satisfy the inequality (make it true). This side is usually shaded to show that it is the correct region,
The ‘boundary line’ will only be a solid line when we have an inequality that involves or . When we have one that uses < or > the line should be drawn as dotted.
|More or less than
|More/less than or equal to
This is used because a point on the line when we have an inequality < or > would not be allowed since the value must be less than or more than another. Therefore, a point on the line which is equal is neither of these things. This can be seen in the way that 3 is not a suitable answer for x in the inequality . However, since any value that is slightly less than 3 is acceptable, we draw a dotted line.
Plot the inequality
To do this we must first convert the inequality by swapping the signs for equals. So we need to plot the line . This is shown below:
Now we must decide if this line should be solid or dotted, and since the inequality has we know that it must stay solid. And also we need to find which part of this line will satisfy the original inequality . We can do this by simply picking any point to one side of the line and if this satisfies the inequality then this side must be shaded; if not then the other side of the line should be. Picking a random point of (2, 0) and filling into the inequality we get or which is clearly correct so we need to shade the area under the line to give the region shown next:
Because the line is solid, we include the points that lie on it and therefore any point that is on the line will satisfy the inequality also. If the line was dashed then this would not be the case and the points that are actually on the line would not satisfy the inequality given, which would have to use ‘less or more than’ signs < or >.
Combining more than one inequality
If we were to be given two different inequality equations then we could put both of these on to a graph and create a region with these which would solve both of the inequalities.
To do this we simply plot the two separately and shade the incorrect regions. Then from this we will be left with a region that will not be shaded and this is the area with points that satisfy both of the stated inequalities.
Plot the following inequalities and mark the region which satisfies both and .
From plotting the correct lines separately for both and and then shading the regions which cannot satisfy the two individually, we are left with a small rectangle in the middle which is not shaded at all. This is the region which satisfies both of the two inequalities.
Producing our own inequalities
Sometimes we may be asked to use real-life situations and convert these into a problem which uses inequalities. By reading a written question carefully and realising what the different unknown values are, this is quite simple.
I hire a car which costs me £50 and then a further 10p for every mile that I cover. If I only have £100 to spend, write down the equality for the miles that I can cover.
First off, we can see that the total amount spent will be equal to where x is the amount of miles we cover. We also need to spend no more than £100, so we can use the inequality
To put this into words we can say that the car must be driven for no more than 500 miles to keep the cost below £100.
Inequalities in two unknowns
So far we have looked at inequalities that only use one unknown but we can also have some which involve two. This involves compromise and you will normally be asked what amounts of each unknown is possible to stick to a total.
A car park is and a car that is parked takes up but a bus takes up . There will be a minimum of 5 buses that turn up to park and there must be at least twice as many cars as there are buses. Plot this problem graphically using inequalities and find the largest number of buses that can be parked.
This obviously sounds very difficult when in words, so we must try to pick out certain points and convert these to inequalities before plotting the problem as a graph.
Let c be the number of car spaces and b be the number of bus spaces.
A car takes up so cars will take up of the car park.
A bus takes up so buses will take up of the car park.
Since the car park is the numbers of cars and buses must satisfy the following:
We can simplify this to:
We are also told that there must be at least 5 buses and the number of cars will be at least twice that of buses. These give us the inequalities:
So we are left with three different inequalities that we can plot on a graph and then find the correct region from:
These are plotted on the next page and the regions which do NOT satisfy each have been shaded accordingly.
The graph above shows the different inequalities as lines with the correctly shaded regions for the parts which do not satisfy them. This then leaves the region which satisfies all of them being unshaded, and the points which have the most number of buses will be to the far right, and has 40 buses and 80 cars, so this is the maximum number that we can have in the car park.
Finding equations from graphs and curves
There are three types of equations that you must be able to interpret and find an equation for from a graph:
1) squared function
2) cubic function
3) exponential function
Each function has two unknowns that need to be found from the graph, a and b. We find these unknowns by reading two different coordinates from the graph and then substituting them into the equations above.
It is much easier if we pick points on either axis as this makes either x or y equal to 0, thus making it easier to work out the values when put into one of the three equations above. Otherwise we will be left with a pair of simultaneous equations to solve.
Use the graph below to find the unknown and in the equation
The graph above clearly passes through the points (0, 2) and (2, 6) and putting these values into the equation for x and y separately will give us the values of a and b.
(0, 2) gives
(2, 6) gives
So the equation which we have is
Doing the same for any of the other equations will give the same effect and will allow you to work out unknown values a and b.
Gradients of straight lines and curves
In an earlier lesson we looked at the gradients of linear equations such as . A positive gradient goes from the bottom left to the top right of the graph and a negative one goes from the top left to the bottom right. Obviously, the steepness may change also.
One of the best ways to find a gradient of a line like this is to picture it as a right-angled triangle and then find the difference in the x value compared with the difference in the y value. The gradient of a line BC is as follows:
It does not matter whereabouts on a line that we do this as the line does not change in gradient from place to place. Therefore, all triangles will produce the same gradient value no matter where on the line they are.
Finding the gradient of a curve by graphing
In an earlier lesson when we looked at circles we learnt that a line that just touches the edge of a circle is known as a tangent:
As well as for a circle, we can also have a tangent to a curve which is also a line that touches the curve just once and does not cross it. A tangent to a curve can be drawn anywhere on the curve and can be used to measure the steepness at a particular point.
The gradient of a curve at a certain point is calculated by drawing a tangent at the point and finding the gradient of this line.
This rule applies to all curves, whether quadratic or otherwise. We simply need to know at what point we wish to find the gradient (since it will clearly change as we move along the curve) and then by drawing a tangent to the curve at this point we will be able to calculate the gradient.
Find the gradient to the curve at the point (2, 0)
From the above graph a tangent to the curve has been placed at the point (2, 0) which can be used to find the gradient by simply finding the gradient of this line. By doing the trick with a right-angled triangle as described above we will find the gradient to be approximately .
This can be done for any curve and any point that is specified. Obviously, placing the tangent and then finding the gradient of the line is done by hand so some human error will occur, but as long as the gradient is approximately correct it will be fine. An examiner will not expect your answer to have a huge amount of accuracy in a question such as this.