Graph each equation and find the point(s) of intersection, if any. The line x+2 y+6=0 and the ci

Graph each equation and find the point(s) of intersection,...

Key Concepts

Substitution Method

The substitution method is an algebraic technique used to solve a system of equations by expressing one variable in terms of another using one equation, and substituting that expression into the other equation(s). This method is particularly useful when dealing with systems where one equation is linear and the other is nonlinear, as it simplifies the process by reducing the problem to a single variable.

System of Equations

A system of equations consists of two or more equations with the same set of unknowns. In algebra and analytic geometry, solving a system means finding the values of the variables that satisfy all equations simultaneously. This approach is essential for determining where graphs, such as those of a line and a circle, intersect.

Graphing of Equations

Graphing equations involves plotting them on a coordinate plane to visually represent the relationships they describe. For linear equations, this results in a straight line, while circles appear as round shapes defined by their center and radius. Graphing aids in understanding the geometric interpretation of solutions and provides an intuitive view of intersections.

Intersection Points

Intersection points are the common solutions where the graphs of two or more equations meet on the coordinate plane. Determining these points involves solving the equations simultaneously, and they represent the set of values that satisfy all the given equations in the system.

Circle Equation in Standard Form

A circle's equation written in standard form, (x - h)² + (y - k)² = r², clearly shows the center (h, k) and the radius r of the circle. This form is fundamental in coordinate geometry because it directly provides the geometric characteristics needed for accurate graphing and analysis of the circle's properties.

Transcript

00:01 For this problem, we're asked to grab the system and find their points of intersection.

00:06 So the first equation is the equation of the circle, so we can identify the center, where when you identify the center, you have to change the sign.

00:16 Negative 1 is positive 1, positive 2 is going to be negative 2.

00:20 So now let's grab our center 1, negative 2.

00:26 And our radius is always, this equation is always equals to r squared, so we know that r has to be 2 because 2 squared is 4.

00:35 So from the center, you're going to go right to, up to, left 2, and down 2.

00:43 So now let's draw a circle that goes through all those four points.

00:52 And for the second equation, that's the equation of a parabola.

00:57 So because, and it's a sideway parabola because y is squared.

01:02 So we need to set this equation equal to x.

01:05 And to do that, you're going to add x to the both sides.

01:09 So therefore, this equation becomes y squared plus 4y plus 1 equals to x.

01:16 So to find the vertex of this side with parabola, you would do the same thing with how you would find the parabola if it was x squared, which is negative b over 2a.

01:27 So that will give you negative 4 over 2a is 1, which is negative 2.

01:33 So to find the x value, you will plug in the negative 2 into the y.

01:44 So negative 2 square is going to give you 4, 4 times negative 2, negative 8 plus 1.

01:51 And 4 minus a is negative 4 plus 1, so that's going to give you negative 3.

01:57 So your vertex for the sideway parabola is located at negative 3 and negative 2.

02:04 So now let's graph that point, negative 3 and negative 2.

02:10 So now the next value you will pick is negative 1.

02:14 So plug in negative 1 into this equation here, and you will see that your y value is going to be a, your x value is going to be a negative 2.

02:24 Same thing with negative 3 is going to be a negative 2.

02:27 And if you plug in 0, your x value is going to be a 1.

02:33 And then if your y value was negative 4, your x value is going to be also 1.

02:40 So now you're going to connect the points, and here's your parabola.

02:47 So your first point of intersection is located at 1 -0, and your second point in intersection located at 1 -9 -4.

02:57 But there seems to be more points of intersection here and here as well.

03:02 So we need to find what those points are exactly by solving this equation.

03:08 So we know what x equals to, so i'm going to substitute that into, this first equation here.

03:14 So x minus 1 squared is going to become y squared plus 4 y plus 1 minus 1 squared and then you copy the rest of them.

03:28 Y plus 2 squared equals to 4.

03:32 So 1 minus 1 is going to be 0.

03:35 So this one we just have y squared plus 4 y squared plus and then foil this so that's going to give you y squared plus 4 y plus 4 equals to 4...

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