Problem 31 - Algebra A Combined Function

Question

Identify whether each equation, when graphed, will be a parabola, circle,ellipse, or hyperbola. Sketch the graph of each equation. If a parabola, label the vertex. If a circle, label the center and note the radius. If an ellipse, label the center. If a hyperbola, label the \(x\) - or \(y\) -intercepts. $$ y=x^{2}+12 x+36 $$

Step-by-Step Solution

Answer

The equation forms a parabola with vertex \((-6, 0)\).

Step 1: Rewrite the Equation

Start by rewriting the equation in a familiar form. Recognize that the equation is quadratic in form: \[ y = x^2 + 12x + 36 \] This resembles the standard quadratic equation \( y = ax^2 + bx + c \). Here, \( a = 1 \), \( b = 12 \), and \( c = 36 \).

Step 2: Identify the Type of Graph

Quadratic equations of the form \( y = ax^2 + bx + c \) graph as parabolas. This is determined by the fact that the highest power of \( x \) is 2, and there is no \( y^2 \) term, which would indicate other conic sections like circles, ellipses, or hyperbolas.

Step 3: Vertex Form Conversion

Complete the square to convert \( y = x^2 + 12x + 36 \) into vertex form. The vertex form of a parabola is given by \( y = a(x-h)^2 + k \), where \((h,k)\) is the vertex. To complete the square:1. Take half of the coefficient of \( x \) (which is 12), square it (resulting in 36), and complete the square:\[ y = (x^2 + 12x + 36) = (x + 6)^2 \]Thus, \( y = (x + 6)^2 \), which simplifies to \( y = (x+6)^2 + 0 \).

Step 4: Identify the Vertex

The vertex can be directly read from the vertex form, \( y = (x+6)^2 + 0 \). The vertex of the parabola is \((-6, 0)\).

Step 5: Sketch the Graph

To sketch the parabola:1. Plot the vertex \((-6, 0)\) on the graph.2. Recognize the parabola opens upwards due to the positive coefficient (1) in \( a(x-h)^2 + k \).3. Draw the symmetrical parabola about the vertical line \( x = -6 \), extending upward.

Key Concepts

Parabola

A parabola is a beautiful, U-shaped curve that provides the graphical representation of a quadratic equation. Parabolas have distinct features:

• The shape is symmetrical, meaning one half is a mirror image of the other.
• The direction it opens (upwards or downwards) is determined by the sign of the coefficient of the squared term.
• Parabolas can also open to the left or right if oriented differently, but standard equations open vertically.

In this exercise, we are dealing with the graph of the quadratic equation. Because of its structure, it can be identified as a parabola opening upwards.

Vertex Form

The vertex form of a quadratic equation is a convenient way to express the equation of a parabola. It is given by:\[ y = a(x - h)^2 + k \]In this form,

• \( h \) and \( k \) represent the coordinates of the vertex of the parabola.
• The parameter \( a \) indicates the direction and width of the opening.

Using vertex form, locating the vertex becomes straightforward. The vertex can tell us the highest or lowest point on the graph, depending on the direction of the parabola. In solving the exercise, converting to vertex form made it simple to identify the vertex as \((-6, 0)\).

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable. The standard form is:\[ y = ax^2 + bx + c \]Here,

• \( a, b, \) and \( c \) are constants,
• \( x \) is the variable.

This form is versatile and allows us to recognize different characteristics of the parabola:

• The value of \( a \) determines the concavity of the parabola.
• By observing \(b\) and \(c\), we infer various properties, such as the position and shape of the parabola.

The exercise starts with identifying equation components, setting the stage for further conversion into vertex form.

Completing the Square

Completing the square is a method used to convert a quadratic equation into its vertex form. This process involves forming a perfect square trinomial from the original equation. Here's a step-by-step summary of completing the square:

• Take the coefficient of the linear term (\(x\)), divide by 2, and square the result.
• Add and subtract this squared value inside the equation to balance it.
• Factor the perfect square trinomial that forms, yielding the equation in vertex form.

In the exercise solution, applying this method to \(y = x^2 + 12x + 36\) led to the vertex form \(y = (x+6)^2\). This form directly revealed the vertex, making graphing intuitive and accessible.