Question

Graph each equation and find the point(s) of intersection, if any. \(y=\frac{4}{x+2}\) and the circle \(x^{2}+4 x+y^{2}-4=0\)

Short answer

There are no points of intersection.

Step-by-step solution

Rewrite the circle equation

The given circle equation is \[x^2 + 4x + y^2 - 4 = 0\].Rewrite it in standard form by completing the square for the quadratic terms in x. Add and subtract 4 inside the equation to complete the square for x: \[x^2 + 4x + 4 + y^2 - 4 = 4\].This simplifies to: \[(x+2)^2 + y^2 = 4\].So the circle equation in standard form is \[(x+2)^2 + y^2 = 4\].

Express y from the first equation

The first equation is given as \[y = \frac{4}{x + 2}\].

Substitute y into the circle equation

Now substitute \[y = \frac{4}{x + 2}\] into the circle equation \[(x+2)^2 + y^2 = 4\].This gives: \[(x+2)^2 + \left(\frac{4}{x+2}\right)^2 = 4\].

Simplify the equation

Expand the squared term in the substituted equation: \[(x+2)^2 + \frac{16}{(x+2)^2} = 4\].Let \[z = (x+2)^2\].This simplifies to: \[z + \frac{16}{z} = 4\].

Solve for z

Multiply every term by z to get rid of the fraction: \[z^2 + 16 = 4z\].Rearrange it into a standard quadratic equation: \[z^2 - 4z + 16 = 0\].

Find the roots

To solve \[z^2 - 4z + 16 = 0\], calculate the discriminant: \[\Delta = b^2 - 4ac = (-4)^2 - 4(1)(16) = 16 - 64 = -48\].Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, there are no real values of z satisfying the equation.

Conclude the solution

Since there are no real solutions for z, it means the equations \[y = \frac{4}{x + 2}\] and \[(x+2)^2 + y^2 = 4\] do not intersect at any real points.

Key concepts

Graphing Equations

Graphing equations helps you visualize how different equations behave across a coordinate system. For example, with \( y = \frac{4}{x+2} \) and a circle \((x+2)^2 + y^2 = 4\), graphing can show their individual shapes and how they may intersect.
To graph, you can plot various points satisfying the equations and connect them smoothly. For \( y = \frac{4}{x+2} \), plot points may reveal a hyperbola opening left and right. The circle will be centered at (-2,0) with a radius of 2. Observing these graphs helps you understand the relationship between the functions and spot intersection points visually.

Quadratic Equations

Quadratic equations typically take the form \( ax^2 + bx + c = 0 \). They can represent parabolas, circles, or hyperbolas based on the variables' relationships.
In our example, we converted the circle's equation into its standard form by completing the square, resulting in \((x+2)^2 + y^2 = 4\). This format provides insights into the shape and position of the circle, crucial for solving intersection points.
Quadratics can be solved using methods like factoring, completing the square, or the quadratic formula, and understanding these solutions helps you handle various complex equations.

Discriminant Analysis

Discriminant analysis involves evaluating the discriminant \( \Delta = b^2 - 4ac \) within a quadratic equation to determine the nature of its roots. A positive \Delta \ indicates two real solutions, zero \Delta \ means one real solution, and a negative \Delta \ implies no real solutions.
In solving \( z^2 - 4z + 16 = 0 \), the discriminant is \(-48\), which is negative. This tells us there are no real solutions, confirming there are no intersection points between the hyperbola and the circle in the real coordinate plane.

Circle Equations

Circle equations in standard form look like \((x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) is the circle's center and \( r \) is its radius.
For our circle \((x+2)^2 + y^2 = 4\), the center is at \((-2, 0)\) with radius 2. Knowing this format simplifies intersections with other equations since we can easily visualize the circle’s properties and its influence on other graphs that share the coordinate system.

Rational Functions

Rational functions are ratios of two polynomials, such as \( y = \frac{4}{x+2} \). They often have vertical asymptotes where the denominator equals zero and horizontal asymptotes describing end behavior.
Here, the function has a vertical asymptote at \( x = -2 \). Plotting points on either side of this asymptote shows how the function behaves. Analyzing its graph helps understand how it intersects (or doesn’t) with other shapes, like our circle, making it a critical aspect of intersections.