As we have seen in our work with inequalities (see this entry to remind yourself if you need), an inequality states a range of solutions to a problem instead of just a singular answer.
The difference is best described with an example:
Here is the line $y=2x+3$y=2x+3
The line shows all the solutions to the equation. All the possible $y$y values that make this equation true for any $x$x value that is chosen.
For every $x$x value there is only one possible corresponding $y$y value.
For example, if $x=1$x=1, then according to the equation $y=5$y=5 (as marked on the diagram)
Here is the inequality $y>2x+3$y>2x+3
The solution to this is not a single line, as for every $x$x value, there are multiple $y$y values that satisfy the inequality. The solution graph is therefore a region.
A coloured in space indicating all the possible coordinates $\left(x,y\right)$(x,y) that satisfy the inequality.
For example, at $x=1$x=1, $y>5$y>5. So any coordinate with an $x$x value of $1$1 and a $y$y value larger than $5$5 is a solution.
The dotted line corresponds to the strictly greater than symbol that was used. That is, since $y$y cannot equal $2x+3$2x+3, we cannot include the points on the line.
Here is another example $y\le2x+3$y≤2x+3
Again we have a region, and this time we also have solid line indicating that the $y$y value can be less than or EQUAL to $2x+3$2x+3, for any given $x$x.
For example, if we choose $x=3$x=3, the points that satisfy the inequality are all the points whose $y$y value is less than or equal to $2\times3+3$2×3+3 or $9$9.
There are many points that do this. One such point would be $\left(3,8\right)$(3,8).
Choose the inequalities that describe the green shaded region.
Which intersection of inequalities describes the shaded region?
The same process applies to non-linear functions.
Again, best demonstrated through an example.
Graph the inequality $y>x^2+1$y>x2+1
Firstly, we need to graph the quadratic $y=x^2+1$y=x2+1. This is a standard quadratic vertically translated up by $1$1 unit, and it looks like this.
Then we look at the inequality statement and interpret it.
$y>x^2+1$y>x2+1, says that we want all $y$y values that are greater then the values created by $y=x^2+1$y=x2+1. So we want the region strictly ABOVE the graph line.
This gives us this graph
Note: the line of the graph is now dotted, because the line itself is not included in the inequality. It would have been if we wanted $y\ge x^2+1$y≥x2+1
Graph the inequality $y\le x^2(x+3)$y≤x2(x+3)
Firstly, we sketch the cubic $y=x^2(x+3)$y=x2(x+3). This cubic has a double root at $x=0$x=0, and a linear root at $x=-3$x=−3. It is positive in nature and so will look like this.
Now, we need to interpret the inequality.
$y\le x^2(x+3)$y≤x2(x+3) means we want all the y values less than OR equal to the cubic curve we have just graphed. So we will include the line (that's the equal to component) and then shade UNDER the line.
So really, graphing inequalities is just one extra step to graphing normal functions. Interpreting the inequality statement and shading in the correct areas.
Consider the system of inequalities defined by $y$y$\ge$≥$x^2$x2 and $y$y$\ge$≥$2x+5$2x+5.
Graph the inequality $y$y$\ge$≥$x^2$x2.
Graph the inequality $y$y$\ge$≥$2x+5$2x+5.
Now, graph the region given by the system $y$y$\ge$≥$x^2$x2 and $y$y$\ge$≥$2x+5$2x+5.
Consider the equations $x^2+y^2=16$x2+y2=16 and $x-y=4$x−y=4.
Graph both equations on the same coordinate plane.
Hence, write down the coordinates of the points of intersection of the two equations. Write the coordinates on the same line, separated by a comma.
Consider the related inequalities $x^2+y^2\ge16$x2+y2≥16 and $x-y>4$x−y>4. Which of the following shaded regions show the solution to the system of inequalities?