Inequalities

Hong Kong

Stage 4 - Stage 5

Lesson

By now, we've covered methods for solving linear and non-linear inequalities and graphing them as regions of the number plane.

If you need to, take your time to revise any of the concepts we've gone through here.

We're now going to use these ideas to write down inequalities for regions that have been described to us in words. Sometimes this will only require a single inequality, but sometimes a system (or group) of several inequalities will need to be listed.

There is a three-step process for constructing inequalities for a given region.

**1)** **BOUNDARIES** - Find the equation(s) for the curve(s) that outline the region.

**2)** **INEQUALITIES** - For each boundary curve, find the inequality that goes with it, paying close attention to the direction of the sign and whether it is strict $<$< or non-strict $\le$≤.

**3)** **CONSTRUCT** - Write the inequality or list the system of inequalities.

Remember, when figuring out the direction of our inequality signs, it is useful to test a point in the specified region by putting it into the equation for the boundary, and observing which way the sign must face.

Write the inequality that satisfies the region below the parabola with vertex $\left(0,25\right)$(0,25), and $x$`x`-intercepts $\left(5,0\right)$(5,0) and $\left(-5,0\right)$(−5,0).

**Boundary**

Given the information, the parabola being described looks like this.

This parabola is the boundary of our region, but what is its equation?

To get this parabola from a standard $y=x^2$`y`=`x`2 parabola, we'd have to first reflect it (flip) and shift it up $25$25 units. This would give us:

$y=-x^2+25$`y`=−`x`2+25

As it turns out, all three of our given points satisfy this equation.

$\left(25\right)=-\left(0\right)^2+25$(25)=−(0)2+25

$\left(0\right)=-\left(-5\right)^2+25$(0)=−(−5)2+25

$\left(0\right)=-\left(5\right)^2+25$(0)=−(5)2+25

Therefore, we need to go no further. $y=-x^2+25$`y`=−`x`2+25 is the equation for the parabola.

We could have also figured this out by using the roots to put the parabola in the form $y=a\left(x-5\right)\left(x+5\right)$`y`=`a`(`x`−5)(`x`+5) and then using the vertex to get $a=-1$`a`=−1.

**Inequality**

Now that we have the boundary, it's time to figure out our inequality.

We want the region 'below' the parabola. Notice that the question didn't say 'below and on' or 'below and including' the parabola, so our inequality is going to be strict $<$<, and points on the curve won't be included, which we indicate with a dotted line.

As mentioned previously, it's good to use a test point to figure out which direction our equality sign will face.

Let's test a point that is well within the region and makes calculation easy. The origin. If we use $\left(0,0\right)$(0,0) for $y=-x^2+25$`y`=−`x`2+25, Then substituting in $y=0$`y`=0 and $x=0$`x`=0 we get $0=-(0)^2+25$0=−(0)2+25 , $0=25$0=25

Now obviously, $0$0 doesn't equal $25$25, so we want to change the $=$= sign to a sign that makes the statement true. $0<25$0<25. So to complete our inequality we get

$y<-x^2+25$`y`<−`x`2+25

A quick way of determining the direction of the inequality sign for functions is to just look at whether the region lies above ($y>f\left(x\right)$`y`>`f`(`x`)) or below ($y`y`<`f`(`x`)) the curve.

**Construct**

We have now constructed our final inequality for the specified region.

$y<-x^2+25$`y`<−`x`2+25

State the system of inequalities describing the region in the first quadrant within the circle centered at the origin with a radius $7$7, and above the line with a gradient $3$3 and a $y$`y`-intercept $-9$−9. (The boundaries are not part of the region.)

Write all inequalities on the same line separated by commas.

**Boundaries**

Let's start with the circle, which looks like this.

This is one of our boundaries, which we will need an equation for. Recall that the equation for a circle centred at the origin with radius $r$`r` has equation $x^2+y^2=r^2$`x`2+`y`2=`r`2. Hence, our equation for this circle boundary will be:

$x^2+y^2=49$`x`2+`y`2=49

The second boundary curve mentioned was the line with gradient $3$3 and $y$`y`-intercept $-9$−9.

We know from our gradient-intercept form of a line $y=mx+b$`y`=`m``x`+`b` (where $m=3$`m`=3 and $b=-9$`b`=−9 in this case) that the above line will have the following equation:

$y=3x-9$`y`=3`x`−9

Last of all, the question specified that the region lay only in the first quadrant. Remember that the first quadrant is the area of the number plane where both $x$`x` and $y$`y` are positive.

Our boundaries for the first quadrant are actually the axes themselves. The $x$`x`-axis is specified by the line $y=0$`y`=0 and the $y$`y`-axis is specified by the line $x=0$`x`=0.

**Inequalities**

Since the first quadrant is the region where both $x$`x` and $y$`y` are positive, it is therefore specified by two inequalities: $x>0$`x`>0 and $y>0$`y`>0. We must also remember that the question specified that the boundaries are not included, so we use strict inequality.

For the circle, the question specified the region inside it.

We can use the equation $x^2+y^2=49$`x`2+`y`2=49 and test the origin for the correct direction of the inequality sign. Clearly $\left(0\right)^2+\left(0\right)^2<49$(0)2+(0)2<49, so we have the following inequality:

$x^2+y^2<49$`x`2+`y`2<49

We could have also worked this out by noticing that the region inside $x^2+y^2=49$`x`2+`y`2=49 is made up of all the circles with a smaller radius than $7$7.

Last of all, for the region above the line, we can recognise that this will be $y>3x-9$`y`>3`x`−9 by using what we discussed earlier about functions and inequalities.

**Construct**

Finally, we list our system of inequalities as follows:

$x>0$`x`>0 , $y>0$`y`>0 , $y>3x-9$`y`>3`x`−9 , $x^2+y^2<49$`x`2+`y`2<49

Write the inequality that satisfies the region inside a circle with a centre at $\left(-6,4\right)$(−6,4) and a radius of $7$7.

State the system of inequalities describing the region in the first quadrant between the lines $y=-x+2$`y`=−`x`+2 and $y=-x+5$`y`=−`x`+5 with the boundaries part of the region.

Write all inequalities on the same line separated by commas.