Problem 5
Question
The graph of each equation is a parabola. Determine whether the parabola opens upward, downward, to the left, or to the right. Do not graph. $$y=-x^{2}+2 x+1$$
Step-by-Step Solution
Verified Answer
The parabola opens downward.
1Step 1: Identify the General Form
The equation given is in the form of a quadratic equation: \( y = ax^2 + bx + c \). Here, \( a = -1 \), \( b = 2 \), and \( c = 1 \).
2Step 2: Determine the Direction of the Parabola
The direction of a parabola (upward or downward) when expressed in the form \( y = ax^2 + bx + c \) is determined by the coefficient \( a \). If \( a > 0 \), the parabola opens upward. If \( a < 0 \), it opens downward. Here, \( a = -1 \) which is less than 0, so the parabola opens downward.
Key Concepts
Parabola DirectionQuadratic GraphingQuadratic Coefficients
Parabola Direction
To determine the direction in which a parabola opens, we must first take a closer look at the quadratic equation in standard form, which is \( y = ax^2 + bx + c \). The key factor here is the coefficient \( a \). This coefficient plays a vital role in dictating the orientation of the parabola's opening.
- If \( a > 0 \), the parabola opens upwards. Imagine it as a U-shaped curve, with the arms reaching towards the sky.
- If \( a < 0 \), like in our equation where \( a = -1 \), the parabola opens downwards. Picture it as an upside-down U, with the curve aiming towards the ground.
Quadratic Graphing
Graphing a quadratic equation can initially feel like a daunting task, but understanding the basic principles can simplify the process significantly. Quadratic graphs are always parabolic, meaning they form a symmetrical "U" shape. This shape can either open upwards or downwards, which is something we determine from the coefficient \( a \) in the equation \( y = ax^2 + bx + c \).
Additionally, identifying specific characteristics of a quadratic graph aids in graphing:
Additionally, identifying specific characteristics of a quadratic graph aids in graphing:
- Vertex: The highest or lowest point on the parabola, depending on its orientation. This is the point where the graph changes direction.
- Axis of Symmetry: A vertical line through the vertex around which the parabola is symmetrical.
- Y-Intercept: The point at which the parabola crosses the y-axis, determined by the constant \( c \) in the equation.
Quadratic Coefficients
The coefficients in a quadratic equation \( y = ax^2 + bx + c \) provide significant information about the parabola's shape and position. Each coefficient has a distinct role:
- Coefficient \( a \): Determines the direction and "width" of the parabola. A larger absolute value of \( a \) makes the parabola "narrower," while a smaller absolute value makes it "wider." For instance, in our example equation, \( a = -1 \), which dictates that the parabola opens downward.
- Coefficient \( b \): Impacts the position of the parabola along the x-axis and affects the symmetry axis's location. It can be said to "tilt" the parabola slightly to the left or right, although it does not affect the width or direction.
- Coefficient \( c \): Represents the y-intercept of the quadratic graph. This is where the parabola will intersect the y-axis \((x = 0)\). In our given equation, \( c = 1 \), meaning the parabola will touch the y-axis at this point.
Other exercises in this chapter
Problem 5
Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{aligned} y^{2} &=4-x \\ x-2 y &=4 \end{aligned}\right. $$
View solution Problem 5
Identify the graph of each equation as an ellipse or a hyperbola. Do not graph. \(-\frac{y^{2}}{25}+\frac{x^{2}}{36}=1\)
View solution Problem 6
Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{l} {x^{2}+y^{2}=4} \\ {x+y=-2} \end{array}\right. $$
View solution Problem 6
Identify the graph of each equation as an ellipse or a hyperbola. Do not graph. \(\frac{y^{2}}{25}+\frac{x^{2}}{36}=1\)
View solution