Problem 82 - Intermediate Algebra

Question

Write each equation in standard form, if it is not already so, and graph it. If the graph is a circle, give the coordinates of its center and its radius. If the graph is a parabola, give the coordinates of its vertex. $$ y=-4(x-4)^{2}-4 $$

Step-by-Step Solution

Answer

This parabola has its vertex at (4, -4) and opens downward.

Step 1: Identify the Type of Equation

The given equation is \( y = -4(x-4)^2 - 4 \). This is in the vertex form \( y = a(x-h)^2 + k \), which describes a parabola. The formula indicates the parabola opens vertically.

Step 2: Determine the Vertex

In the vertex form \( y = a(x-h)^2 + k \), the vertex is located at \((h, k)\). From \( y = -4(x-4)^2 - 4 \), we identify \( h = 4 \) and \( k = -4 \), giving a vertex at \((4, -4)\).

Step 3: Convert to Standard Form

The standard form of a parabola equation is \( y = ax^2 + bx + c \). Expand and simplify the given equation: \( y = -4(x^2 - 8x + 16) - 4 \). By distributing the \(-4\), we have \( y = -4x^2 + 32x - 64 - 4 \) which simplifies to \( y = -4x^2 + 32x - 68 \).

Step 4: Graph the Equation

Plot the vertex at \((4, -4)\). Since \( a = -4 \), the parabola opens downward and is relatively narrow. Sketch the parabola reflecting a downward arc centered on the vertex.

Key Concepts

Vertex Form

The vertex form of a quadratic equation is incredibly useful for easily identifying key characteristics of a parabola. In vertex form, a quadratic equation appears as \( y = a(x - h)^2 + k \). Here, the values of \( h \) and \( k \) reveal significant information:

• The vertex of the parabola is at \( (h, k) \).
• The value of \( a \) tells us about the direction and width of the parabola.
• A positive \( a \) value implies the parabola opens upwards, whereas a negative \( a \) means it opens downwards.
• The larger the absolute value of \( a \), the narrower the parabola, while a smaller absolute value indicates a wider parabola.

In the example equation, \( y = -4(x-4)^2 - 4 \), the vertex is at \( (4, -4) \) and the downward direction is due to \( a = -4 \). This is why understanding the vertex form is essential, as it simplifies the task of determining the vertex and the parabola's orientation.

Standard Form

Standard form is another way to express a quadratic equation, which is generally structured as \( y = ax^2 + bx + c \). This form can provide different insights into the equation compared to vertex form.

• It is useful for identifying the y-intercept, which occurs at \( (0, c) \).
• It helps in finding the roots or x-intercepts of the parabola through methods like factoring or the quadratic formula.
• It’s easier to identify the overall symmetry of the parabola using this form.

In our example, the equation \( y = -4(x-4)^2 - 4 \) was transformed into standard form: \( y = -4x^2 + 32x - 68 \). This linear arrangement clearly shows how coefficients \( a \), \( b \), and \( c \) interact to shape the parabola.

Graphing Parabolas

Graphing a parabola involves several steps, and understanding both the vertex and standard forms can enhance this process. Begin by plotting the vertex, as it serves as a central point of the graph. For our equation, the vertex is \( (4, -4) \).
Next, analyze the value of \( a \). Since \( a = -4 \) in our equation, the parabola opens downwards, so it’s essential to reflect this in the graph by drawing a curve that descends from the vertex. Additionally, the steep magnitude of \( 4 \) suggests that the parabola will be relatively narrow.
To improve accuracy further, identify a few points on the graph:

• Choose x-values around the vertex and plug them into either the vertex or standard form to find their corresponding y-values.
• Calculate the x-intercepts by setting \( y = 0 \) and solving for \( x \).
• Use symmetry, as a parabola is symmetric about the vertical line through its vertex.

Graphing becomes easier as these various methods and clues combine, providing a comprehensive picture of the parabola's trajectory.