Problem 81 - 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. $$ y=-4(x+5)^{2}+5 $$

Step-by-Step Solution

Answer

The vertex of the parabola is at \((-5, 5)\). The parabola opens downward.

Step 1: Identify the Equation Type

The given equation is \( y = -4(x + 5)^2 + 5 \). It is in the general form of a parabola \( y = a(x-h)^2 + k \), where \( a \), \( h \), and \( k \) are constants.

Step 2: Graph the Parabola

Since the equation is in the form \( y = a(x-h)^2 + k \), this represents a parabola that opens upward if \( a > 0 \) and downward if \( a < 0 \). Here, \( a = -4 \), so it opens downwards. The vertex of the parabola is at \((h, k)\).

Step 3: Find the Vertex

In the equation \( y = -4(x + 5)^2 + 5 \), we identify \( h = -5 \) and \( k = 5 \). Therefore, the vertex of the parabola is \((-5, 5)\).

Step 4: Write the Equation in Standard Form

The standard form of a parabola that opens vertically is \( x^2 = 4py \) with vertex \((h, k)\). However, this equation is already suitable for graphing as it stands, in vertex form. There is no need for further conversion for graphing purposes.

Key Concepts

Vertex Form of a Quadratic Equation

The vertex form of a quadratic equation is a special format of expressing a quadratic function. It is given by the formula: \( y = a(x-h)^2 + k \), where:

• \(a\) represents the stretch or compression factor of the graph and determines the direction of the parabola’s opening.
• \(h\) and \(k\) are the coordinates of the vertex, \( (h, k) \).

The vertex form is useful because it directly provides the vertex of the parabola. This makes it easier to identify the position of the parabola on the graph without further calculations. In our example, the equation \( y = -4(x + 5)^2 + 5 \) is perfectly written in vertex form, indicating that the parabola opens downward due to the negative \(a\)-value of \(-4\). The vertex here is \((-5, 5)\), which gives the highest point on the graph for this downward-opening shape.

Graphing Quadratic Equations in Vertex Form

Graphing a quadratic equation that is in vertex form can be straightforward once you understand its components. Here's how you can do it:

• Identify the vertex from the equation, \((h, k)\). This is the starting point for your graph.
• Determine the direction of the parabola. If \(a > 0\), it opens upwards, and if \(a < 0\), it opens downwards.
• Plot the vertex on a coordinate plane.
• From the vertex, use the \(a\) value to find additional points. For instance, since \(a\) affects the width of the parabola, this helps you find points equidistant from the vertex.
• Draw the axis of symmetry, a vertical line through the vertex, which helps in plotting symmetric points.

Understanding these steps lets you quickly sketch the parabola without needing to convert the equation into standard form.

Vertex of a Parabola

The vertex of a parabola is the point where the curve reaches a maximum or minimum value. It is a key feature when graphing quadratic equations and provides essential information about the shape and position of the parabola.
For a quadratic equation in vertex form \( y = a(x-h)^2 + k \), the vertex is given by the coordinates \((h, k)\). This means the vertex acts as a guidepost on the graphing journey, helping to define symmetry and the extremities of the parabola.
In our particular equation, \( y = -4(x + 5)^2 + 5 \), the vertex is \((-5, 5)\). This indicates the highest point of the parabola because \(a\) is negative, causing it to open downward. By knowing the vertex, you can easily plot this pivotal point and draw the rest of the parabola accordingly, ensuring an accurate graph.