Week 11 – Precalc 11

One of the things we learned this week in Precalc 11 is how to graph Quadratic Inequalities in Two Variable (x and y).

Graph the inequality.

y < x^2 + 8x +7

The inequality is in general form so first step in graphing this inequality is converting it to standard/vertex form by completing the square.

y < x^2 + 8x +7
y < x^2 + 8x + 16 - 16 + 7
y < (x+4)^2 - 9

Now, we can determine the coordinates of the vertex and graph the parabola.

Next, we have to determine which part of the graph is shaded, to do that, we have to test points inside or outside the parabola and whatever side makes the inequality true, we shade it in.

Test point (0, 0)
0 < 0^2 + 8(0) +7
0 < 7
The origin is part of the solution so outside of the parabola should be shaded and since its less than not less than or equal to the line should be dashed lines. Let’s try a coordinate inside the parabola.
Test point (-5, 0)
0 < (-5)^2 + 8(-5) + 7
0 < 25 - 40 + 7
0 < -8
-8 is not greater than 0 so coordinates inside the parabola are not solutions.

Write the inequality for the graph

 

The first step is to find the coordinates vertex which is (5, -3). Next determine the stretch or compression pattern which is 2-6-10, with these information we can fill make a standard/vertex form equation. y = -2(x - 5)^2 -3 so now we have to figure out whether its <, >, \leq or \geq by testing coordinates. Testing coordinates inside the shaded area would be better because coordinates in the non-shaded area are not solutions so you will have to flip the sign afterwards to make it a false statement.

Test point (0, 0)

0 = -2(0 - 5)^2 - 3

 

0 = -2(25) - 3

 

0 = -53

Since (0, 0) is part of the solution, we have to make the statement true and since the line of the parabola is solid, it’s either \leq or \geq

0 \geq-53
y \geq -2(x - 5)^2 -3

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