Solving quadratic inequalities graphically
Now, what if we are given an inequality involving a quadratic, such as $x^2+2x-8>0$x2+2x−8>0? We do not need to know an algebraic method for quadratic inequalities in this course and the graphical method provides an efficient way to find our solutions. We can consider the quadratic function $y=x^2+2x-8$y=x2+2x−8 and find values of $x$x where the function is greater than $0$0.
Let's use what we already know about quadratics to sketch a graph of $y=x^2+2x-8$y=x2+2x−8. We know that the parabola will be concave up, since the coefficient of $x^2$x2 is positive. We also get a $y$y-intercept of $y=-8$y=−8 after substituting $x=0$x=0. We can use our vertex formula $x=\frac{-b}{2a}$x=−b2a to find the axis of symmetry where our turning point lies, which we know will be a minimum turning point due to the concavity.
| $x$x | $=$= | $\frac{-b}{2a}$−b2a |
| $=$= | $\frac{-2}{2\times1}$−22×1 | |
| $=$= | $-1$−1 |
Last of all, we can also factorise the quadratic to find our $x$x-intercepts.
| $x^2+2x-8$x2+2x−8 | $=$= | $0$0 |
| $\left(x+4\right)\left(x-2\right)$(x+4)(x−2) | $=$= | $0$0 |
| $x$x | $=$= | $-4$−4, $2$2 |
Now we can sketch a graph of our parabola.
Being asked to find solutions to $x^2+2x-8>0$x2+2x−8>0 is exactly the same as being asked to find values of $x$x at which the graph of $y=x^2+2x-8$y=x2+2x−8 is above the line $y=0$y=0 (the $x$x-axis). In other words, when is $y=x^2+2x-8$y=x2+2x−8 positive?
We can see from the graph here that unlike our previous inequality which provided us with a single range of values for our answer, here, there are two. Our answer has to be written in two parts $x<-4$x<−4 and $x>2$x>2. In interval notation, this could be written as $(-\infty,-4)\cup(2,\infty)$(−∞,−4)∪(2,∞).
What if we had flipped the inequality sign in the question? That is, we were required to solve $x^2+2x-8<0$x2+2x−8<0. Now we are being asked when the graph $y=x^2+2x-8$y=x2+2x−8 is negative. Now our solutions will be $x>-4$x>−4 and $x<2$x<2. This can be written as $-4
Equivalent inequalities
Suppose we want to solve $\left(x+2\right)^2>2x+12$(x+2)2>2x+12. One approach would be to go through the slightly complicated task of sketching graphs of the parabola on the left and the line on the right to examine when the parabola lies above the line. These graphs would have to be quite accurate to ensure we can clearly identify the points of intersection.
However, we could simplify the inequality first, bringing all of the terms to the left-hand side so that we only need to sketch one graph. Expanding and simplifying, we get:
| $\left(x+2\right)^2$(x+2)2 | $>$> | $2x+12$2x+12 |
| $x^2+4x+4$x2+4x+4 | $>$> | $2x+12$2x+12 |
| $x^2+2x-8$x2+2x−8 | $>$> | $0$0 |
The inequality $\left(x+2\right)^2>2x+12$(x+2)2>2x+12 is thus equivalent to the inequality $x^2+2x-8>0$x2+2x−8>0 from before, and so it will have the same solutions.
Practice questions
Question 1
Consider the graph of $y=f\left(x\right)$y=f(x).
Find the values of $x$x for which $f\left(x\right)=0$f(x)=0. Write both solutions on the same line separated by a comma.
For what values of $x$x is $f\left(x\right)<0$f(x)<0?
Give your answer as an inequality.
For what values of $x$x is $f\left(x\right)>0$f(x)>0?
Give your answer as an inequality.
What is the $x$x-coordinate of the vertex of $f\left(x\right)$f(x)?
Question 2
We want to solve $x^2-2x\le-x+2$x2−2x≤−x+2 by plotting two curves and finding their points of intersection.
First find the $x$x-intercepts of $y=x^2-2x$y=x2−2x.
Find the $y$y-intercept of $y=x^2-2x$y=x2−2x.
Find the $x$x-intercept of $y=-x+2$y=−x+2.
Find the $y$y-intercept of $y=-x+2$y=−x+2.
Hence plot the set of curves for $y=x^2-2x$y=x2−2x and $y=-x+2$y=−x+2.
Loading Graph...Use the graph from the previous part to select the correct option:
The two curves intersect at $\left(-1,3\right)$(−1,3) and $\left(2,0\right)$(2,0), hence the solution to the inequality is$-1\le x\le2$−1≤x≤2.
AThe two curves intersect at $\left(-1,3\right)$(−1,3) and $\left(2,0\right)$(2,0), hence the solution to the inequality is $x\le-1,x\ge2$x≤−1,x≥2.
BThe two curves intersect at $\left(3,-1\right)$(3,−1) and $\left(0,2\right)$(0,2), hence the solution to the inequality is $0\le x\le3$0≤x≤3.
CThe two curves intersect at $\left(3,-1\right)$(3,−1) and $\left(0,2\right)$(0,2), hence the solution to the inequality is $x\le0,x\ge3$x≤0,x≥3.
D