Solving Algebraic Problems with Graphs

Question

I've come across this problem in my textbook:

solve y < 0

According to the textbook, I have to use a graph on the page to solve the problem. The function of the graph is y = x^2 - 4 and so the two values on the x-axis are -2 and 2.

According to the textbook the answer is -2 < x < 2 but it doesn't explain how they got that answer.

The following problem wants you to solve y > 0 (using the same graph function) and the answer is x < -2 or x > 2 but like the previous example, they don't explain how they came up with that solution.

I'd be thankful if someone could explain this to me.

Answer 1

Hint: for y<0, simply try putting x as a value between 2 and -2? U should automatically figure out what's hapenning. Or simply plot the graph of $x^2-4$ to see the -2 to +2 region in 4 th quadrant.

Answer 2

You are looking for when $x^2-4>0$.

If you look at the graph, it is a large $u$ that is mostly above the x-axis, but it is under the x-axis for some range of values.

These places where the function hits the x-axis, they are called roots .

The roots of $x^2-4$ are -2 and 2, because they are the solutions to the equation $x^2-4=0$.

Between these roots, $x^2-4<0$, at these roots $x^2-4=0$, and otherwise, $x^2-4>0$.

Answer 3

Once you have graphed the function you can easily see for which $x$ the inequality $y < 0$ holds. You just have to look where the graph runs below the $x$-axis. As you can surely see, this is exactly the region $-2 < x < 2$ since the graph intersects with the $x$-axis at $x=2$ and $x=-2$ and between these two points it runs below the line $y=0$, i. e. the $x$-axis. The same argumentation holds for the case $y>0$.

Answer 4

The red portion of the following graph of $y=x^2-4$ is the portion for which the $y$-coordinates are negative.

The question pertains only to those points of the graph: Over what interval do the $x$ values of those points range?

Looking at the graph we see that the $x$ coordinates of the red points must lie on the interval $(-2,2)$.

Notice that $-2$ and $2$ are not included in the interval because the $y$ coordinate is not less than zero there, rather it is equal to zero there.