Problem 114
Question
Parabolas \begin{equation}\begin{array}{l}{\text { a. Find the coordinates of the vertex of the parabola }} \\ \quad {y=a x^{2}+b x+c, a \neq 0}. \\ {\text { b. When is the parabola concave up? Concave down? Give }} \\ \quad{\text { reasons for your answers. }}\end{array}\end{equation}
Step-by-Step Solution
Verified Answer
The vertex of the parabola is at \((-\frac{b}{2a}, y)\). It is concave up if \(a > 0\), and concave down if \(a < 0\).
1Step 1: Formula for the Vertex
The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \( x = -\frac{b}{2a} \).
2Step 2: Find the x-coordinate of the vertex
Substitute the values of \( a \) and \( b \) from the equation \( y = ax^2 + bx + c \) into the formula: \( x = -\frac{b}{2a} \). This calculates the x-coordinate of the vertex.
3Step 3: Find the y-coordinate of the vertex
Substitute the x-coordinate of the vertex back into the original parabola equation \( y = ax^2 + bx + c \) to find the corresponding y-coordinate, thus obtaining the coordinates of the vertex \((x, y)\).
4Step 4: Determine concavity
The parabola is concave up if \( a > 0 \) and concave down if \( a < 0 \), because the sign of \( a \) determines the direction of the opening of the parabola. If \( a > 0 \), the arms of the parabola open upwards, and if \( a < 0 \), they open downwards.
Key Concepts
Vertex of a ParabolaConcavity of a ParabolaQuadratic Equationsx-coordinate of the Vertex
Vertex of a Parabola
The vertex of a parabola is a crucial point, as it represents the peak or the lowest point of the "U" shape formed by the parabola. In a quadratic equation of the form \( y = ax^2 + bx + c \), the vertex can be found using a very straightforward method.
First, you determine the x-coordinate of the vertex by utilizing the formula \( x = -\frac{b}{2a} \). This formula comes from completing the square of the quadratic equation and provides a way to find where the axis of symmetry of the parabola lies.
Once you have the x-coordinate, substitute this value back into the original quadratic equation \( y = ax^2 + bx + c \) to solve for the y-coordinate. This gives you the full coordinates of the vertex \((x, y)\), pinpointing this critical spot on the graph of the parabola.
First, you determine the x-coordinate of the vertex by utilizing the formula \( x = -\frac{b}{2a} \). This formula comes from completing the square of the quadratic equation and provides a way to find where the axis of symmetry of the parabola lies.
Once you have the x-coordinate, substitute this value back into the original quadratic equation \( y = ax^2 + bx + c \) to solve for the y-coordinate. This gives you the full coordinates of the vertex \((x, y)\), pinpointing this critical spot on the graph of the parabola.
Concavity of a Parabola
Understanding the concavity of a parabola is essential to figuring out how it behaves visually. When we talk about concavity, we're referring to whether the parabola opens upwards or downwards. This is determined by the sign in front of the \( a \) term in the equation \( y = ax^2 + bx + c \).
If \( a > 0 \), the parabola opens upward. This means it looks like a cup or a bowl, with the vertex being the lowest point.
If \( a < 0 \), the parabola opens downward. In this orientation, it resembles an upside-down bowl, and the vertex is the highest point.
If \( a > 0 \), the parabola opens upward. This means it looks like a cup or a bowl, with the vertex being the lowest point.
If \( a < 0 \), the parabola opens downward. In this orientation, it resembles an upside-down bowl, and the vertex is the highest point.
- This distinction is important in various fields, such as physics and economics, where the direction of a curve could imply maximum or minimum values.
Quadratic Equations
Quadratic equations are a type of polynomial equation involving a square term. They're typically expressed in the standard form \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero.
Quadratic equations are particularly known for their use in modelling scenarios involving projectile motion or problems concerning area, revenue, and cost in business.
One key feature of quadratic equations is that they graph as parabolas. This allows for analysis of important properties, such as the vertex and roots (where the parabola intersects the x-axis).
Quadratic equations are particularly known for their use in modelling scenarios involving projectile motion or problems concerning area, revenue, and cost in business.
One key feature of quadratic equations is that they graph as parabolas. This allows for analysis of important properties, such as the vertex and roots (where the parabola intersects the x-axis).
- This type of equation also allows us to calculate various outputs given any input, which is particularly useful in a broad range of applications.
x-coordinate of the Vertex
The x-coordinate of the vertex is derived from the quadratic formula and plays a vital role in determining the position of the vertex on the graph of a parabola.
This is calculated using \( x = -\frac{b}{2a} \), where \( b \) and \( a \) are coefficients from the quadratic equation \( y = ax^2 + bx + c \). This formula is essentially the formula for the axis of symmetry for the parabola.
This is calculated using \( x = -\frac{b}{2a} \), where \( b \) and \( a \) are coefficients from the quadratic equation \( y = ax^2 + bx + c \). This formula is essentially the formula for the axis of symmetry for the parabola.
- Knowing the x-coordinate helps in sketching the graph of the parabola and understanding where it narrows or widens. This part of the calculation is particularly helpful in various mathematical and real-world situations where finding the optimal point is necessary.
Other exercises in this chapter
Problem 113
If \(b, c,\) and \(d\) are constants, for what value of \(b\) will the curve \(y=x^{3}+b x^{2}+c x+d\) have a point of inflection at \(x=1 ?\) Give reasons for
View solution Problem 113
Use a CAS to solve the initial value problems in Exercises \(111-114 .\) Plot the solution curves. $$ y^{\prime}=\frac{1}{\sqrt{4-x^{2}}}, \quad y(0)=2 $$
View solution Problem 114
Use a CAS to solve the initial value problems in Exercises \(111-114 .\) Plot the solution curves. $$ y^{\prime \prime}=\frac{2}{x}+\sqrt{x}, \quad y(1)=0, \qua
View solution Problem 115
Quadratic curves What can you say about the inflection points of a quadratic curve \(y=a x^{2}+b x+c, a \neq 0 ?\) Give reasons for your answer.
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