Question Video: Solve a Quadratic Inequality by Graphing | Nagwa Question Video: Solve a Quadratic Inequality by Graphing | Nagwa

Question Video: Solve a Quadratic Inequality by Graphing Mathematics

Solve βˆ’π‘₯Β² + 9 > 0 graphically.

04:11

Video Transcript

Solve negative π‘₯ squared plus nine is greater than zero graphically.

Negative π‘₯ squared plus nine is greater than zero is a quadratic inequality. In order for us to graph that, let’s consider the general form of a quadratic equation. 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐. When we have a general form of a quadratic, we know a few of its features that help us graph the quadratic. We can find the shape of the parabola. When the π‘Ž-value is greater than zero, the parabola opens upward. And when the π‘Ž-value is less than zero, the parabola opens downward. When we’re given this general form, the 𝑦-intercept is located at zero, 𝑐. And we have roots, the places where the graph crosses the π‘₯-axis when 𝑓 of π‘₯ equals zero.

To solve the inequality, we’ll first solve the equation 𝑓 of π‘₯ equals negative π‘₯ squared plus nine for these three features. In this case, the π‘Ž-value is negative one. And we know our graph will open downward. The 𝑦-intercept will be at zero, nine. And to find the roots, the place where this graph crosses the π‘₯-axis, we’ll set negative π‘₯ squared plus nine equal to zero. To solve for π‘₯, we want it by itself. So we’ll add π‘₯ squared to both sides. Then we have π‘₯ squared equals nine. If we take the square root of both sides of the equation, we see that π‘₯ equals plus or minus three. And that means when π‘₯ is positive three, the graph crosses the π‘₯-axis. And when π‘₯ is negative three, the graph crosses the π‘₯-axis.

Using these three points and the fact that our graph opens downward, we can start to graph this inequality. Once we’ve sketched our coordinate plane, we can graph the 𝑦-intercept at zero, nine and the roots at negative three, zero and three, zero. And this is where we need to start making some distinctions. While we used 𝑓 of π‘₯ equals negative π‘₯ squared plus nine to find the roots and the shape, the graph of the inequality will look a bit different than if we were graphing a quadratic equation. Because we’re dealing with the function that is greater than but not equal to, we’ll connect the points with a dotted parabola.

In addition to that, we’ll want to shade the area of the graph that makes this inequality true, the places where negative π‘₯ squared plus nine is greater than zero. This is the space on the graph above the π‘₯-axis, the space where the 𝑦-values are always greater than zero. This image is the graph of the quadratic inequality negative π‘₯ squared plus nine is greater than zero. However, when it comes to solving this, that means we’re looking for the range of π‘₯-values that makes the statement true. We want to say, for what π‘₯-values is negative π‘₯ squared plus nine greater than zero?

On our graph, we see that this happens between the two roots. So from negative three to positive three π‘₯-values, this function is greater than zero. We want to write this in set notation, which is negative three to three. We use parentheses to indicate that negative three does not make the statement true nor does positive three make the statement true. If we plug in negative three for π‘₯, the outcome is zero. And zero is not greater than zero. This means we want to exclude negative three and positive three and only include the values between negative three and positive three. Another form you might see it written in as brackets facing outward. Both of these are ways to show that negative three and positive three are not included, only the values between.

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