Problem 46 - Contemporary Precalculus

Question

In Exercises \(43-54\), find the equation of the parabola satisfying the given conditions. Vertex (-3,0)\(;\) axis \(y=0 ;(-1,1)\) on graph.

Step-by-Step Solution

Answer

**Short answer:** The equation of the parabola with vertex (-3, 0), axis y = 0, and point (-1, 1) on the graph is \((y)^2 = \frac{1}{2}(x + 3)\).

Step 1: Insert the vertex coordinates into the equation

Plug in the vertex (-3, 0) into the standard form of the parabola equation: \((y - k)^2 = 4a(x - h)\). Since the vertex is (-3, 0), we have (h, k) = (-3, 0). The equation becomes: \((y - 0)^2 = 4a(x - (-3))\) or \((y)^2 = 4a(x +3)\).

Step 2: Plug in the given point into the equation

Now, we will insert the given point (-1, 1) into the equation and solve for "a". \((1)^2 = 4a((-1) + 3)\) => \(1 = 4a(2)\) => \(1 = 8a\)

Step 3: Solve for "a"

Divide both sides of the equation by 8 to isolate "a": \(1 = 8a\) \(a= \frac{1}{8}\)

Step 4: Write the equation of the parabola

Now that we have found the value of "a", we can write the equation of the parabola by substituting the value of "a" into the equation \((y)^2 = 4a(x + 3)\). \((y)^2 = 4(\frac{1}{8})(x + 3)\) Simplify the equation by multiplying 4 with \(\frac{1}{8}\): \((y)^2 = \frac{1}{2}(x + 3)\) This is the equation of the parabola that satisfies the given conditions: vertex (-3,0), axis y=0, and (-1,1) on the graph.

Key Concepts

Vertex

The vertex of a parabola is a crucial point that helps in understanding its structure and direction. It is essentially the peak or the lowest point of the parabola, depending on its orientation. For our specific problem, the vertex is given as (-3, 0).

The coordinates of the vertex are used in the standard form of the parabola's equation, \((y - k)^2 = 4a(x - h)\), where (h, k) are the coordinates of the vertex. Knowing the vertex provides essential information about how the parabola is positioned on the graph.

• The vertex form helps in easily identifying the vertex from the equation.
• The vertex indicates if the parabola is opening upward, downward, left or right.
• This point is also known as the turning point.

With the vertex at (-3, 0), the parabola is directed along the x-axis, as confirmed by the standard form equation \((y)^2 = (x + 3)\).

Axis of Symmetry

The axis of symmetry is a line that splits the parabola into two mirror-image halves. Every parabola has a vertical or horizontal axis of symmetry, depending on its orientation. The axis can be determined quickly once the vertex position is known.

For the given exercise, the axis of symmetry is along the line y = 0, which means the parabola is oriented horizontally.

• A horizontal axis of symmetry suggests that the parabola opens left or right.
• The axis of symmetry always passes through the vertex.
• It is a helpful reference line for sketching the parabola accurately.

Knowing the axis of symmetry aids in predicting the shape and orientation, which are key to graphing the parabola on any coordinate plane.

Standard Form

The standard form of a parabola provides a structured way to identify and write the equation of the parabola. The general equation in standard form is \((y-k)^2 = 4a(x-h)\). This form highlights the role of the vertex, as (h, k) provides its coordinates.

Once you insert the vertex into this form, you can determine the orientation by understanding the value of "a."

• The standard form highlights the parabola's symmetry.
• By substituting known points, it's possible to solve for "a" and fully determine the parabola's equation.
• In this problem, solving for "a" allows us to refine the equation to \((y)^2 = \frac{1}{2}(x + 3)\),which fully characterizes the parabola based on the given conditions.

Using the standard form is an efficient process to determine and express how a parabola behaves geometrically.

Coordinate Geometry

Coordinate geometry is a fundamental area of mathematics that deals with shapes like parabolas through algebraic equations. It's all about representing geometric shapes in a coordinate system, which provides extensive analytical insights.

In our exercise, coordinate geometry is employed to frame and solve the equation of the parabola using provided points and conditions.

• This technique integrates algebra with geometry for precise curve plotting.
• It ensures descriptions and transformations of shapes are carried out mathematically.
• The process involves setting conditions like a vertex and axis, and it reveals the curve's orientation and position.

By using coordinate geometry, one can visualize and solve problems related to parabolic curves efficiently, as it forms a bridge between theoretical math and real-world problems.