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Magazine Chapter 2: Problem 24
In exercises \(21-24,\) expand and graph the function. $$ x(y)=(y+2)^{2}-1 $$
Short Answer
Expert verified
The expanded form is \(x(y) = y^2 + 4y + 3\). The graph is a parabola with vertex \((-2, -1)\) and passes through points \((-3, 0), (-1, 0), (0, 3)\).
Step by step solution
01
Expand the Expression
First, we need to expand the expression \((y+2)^2 - 1\). To do this, we will expand \((y+2)^2\). Apply the formula for a binomial square: \((a+b)^2 = a^2 + 2ab + b^2\). Here, \(a = y\) and \(b = 2\), so \((y+2)^2 = y^2 + 4y + 4\).
02
Simplify the Expanded Expression
Now simplify by subtracting 1 from the expanded binomial. Substitute \(y^2 + 4y + 4\) back into the expression: \((y+2)^2 - 1 = y^2 + 4y + 4 - 1 = y^2 + 4y + 3\). This is the expanded form of the function \(x(y)\).
03
Identify Key Features of the Function
The function \(x(y) = y^2 + 4y + 3\) is a quadratic function in terms of \(y\) and will form a parabola when graphed. Identify the vertex using the formula \(y = -\frac{b}{2a}\) for the vertex in \(y\). Here, \(a = 1\) and \(b = 4\), so \(y = -\frac{4}{2\times1} = -2\). Substitute \(y = -2\) back into the expression to find \(x\): \(x(-2) = (-2)^2 + 4(-2) + 3 = 4 - 8 + 3 = -1\). The vertex is \((-2, -1)\).
04
Determine Additional Points
Choose values of \(y\) near the vertex to find additional points. Let's use \(y = -3, -1, 0\). For \(y = -3\), \(x(-3) = (-3)^2 + 4(-3) + 3 = 9 - 12 + 3 = 0\). For \(y = -1\), \(x(-1) = (-1)^2 + 4(-1) + 3= 1 - 4 + 3 = 0\). For \(y = 0\), \(x(0) = 0^2 + 4(0) + 3 = 3\). Thus, additional points are \((-3, 0), (-1, 0), (0, 3)\).
05
Graph the Function
Plot the vertex \((-2, -1)\) and the additional points \((-3, 0), (-1, 0), (0, 3)\) on a Cartesian coordinate plane. The parabola opens upwards because the coefficient of \(y^2\) is positive. Draw a smooth curve through the points to graph the function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola
In mathematics, a parabola is a symmetric curve formed by graphing a quadratic equation. It represents the path traced by a point at a fixed distance from a fixed point called the "focus" and a fixed line called the "directrix." This type of curve is prevalent in various real-world applications, such as the trajectory of projectiles.
- A parabola is characterized by its U-shape, which can open upward or downward.
- The direction of the opening is determined by the sign of the coefficient of the squared term in the quadratic equation.
- If the coefficient is positive, the parabola opens upwards; if negative, it opens downwards.
Vertex
The vertex is a significant point on a parabola, representing the maximum or minimum value of the quadratic function depending on its orientation.
- It's located at the axis of symmetry, which splits the parabola into mirror images.
- The vertex can provide valuable insights into the characteristics of the function, such as its peak for a downward opening parabola or dip for an upward opening one.
Binomial Expansion
Binomial Expansion is a method to expand expressions that are raised to a power, written in the form \((a+b)^n\). For quadratic functions, this expansion plays a crucial role in writing and simplifying expressions. Here's how you can engage with this concept:
- Apply the Binomial Theorem: \((a+b)^2 = a^2 + 2ab + b^2\).
- This allows us to rewrite quadratic functions in a more manageable way.
- For example, expanding \((y+2)^2\) gives us \(y^2 + 4y + 4\). This helps in understanding and solving the equation.
Graphing Quadratics
Graphing quadratic functions involves plotting a set of points that satisfy the function, usually forming a parabola. It's a visual approach to understand the behavior of quadratic equations. Here are some steps and tips for graphing:
- The first step is expanding and simplifying the quadratic function, making it easier to identify key components.
- Finding the vertex provides a starting point, indicating where the curve has its lowest or highest point.
- Next, calculate additional points by substituting values near the vertex, providing a fuller shape of the curve.
