Chapter 0: Problem 19
Graph. \(y=-x^{2}+2 x-1\)
Short Answer
Expert verified
The parabola opens downward, has vertex (1, 0), y-intercept (0, -1), and axis of symmetry x = 1.
Step by step solution
01
Identify the Type of Function
The equation given is a quadratic function, which can be written in the form of \[y = ax^2 + bx + c\]where \(a = -1\), \(b = 2\), and \(c = -1\). Quadratic functions lead to parabolic graphs.
02
Determine the Direction of the Parabola
Since the coefficient \(a = -1\) is negative, the parabola opens downwards. This is because a negative leading coefficient in a quadratic equation results in a downward-opening parabola.
03
Find the Vertex
The vertex form of a quadratic is \[y = a(x-h)^2 + k\] where \((h, k)\) is the vertex. The vertex can also be found using \[h = -\frac{b}{2a}\].For this equation, \[h = -\frac{2}{2(-1)} = 1\].Substitute \(x = 1\) into the equation to find \(k\):\[y = -(1)^2 + 2(1) - 1 = 0\]. Hence, the vertex is \((1, 0)\).
04
Find the Axis of Symmetry
The axis of symmetry for the parabola can be found using the x-coordinate of the vertex. Thus, the axis of symmetry is \[x = 1\].
05
Calculate the Y-Intercept
The y-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the equation: \[y = -(0)^2 + 2(0) - 1 = -1\].Therefore, the y-intercept is at \((0, -1)\).
06
Determine Additional Points
Select a few x-values around the vertex to find points through which the parabola passes. For example, if \(x = 2\): \[y = -(2)^2 + 2(2) - 1 = -1\]. Therefore, the point \((2, -1)\) is on the graph. Also, for \(x = -1\):\[y = -(-1)^2 + 2(-1) - 1 = -4\]; hence, point \((-1, -4)\) is also on the graph.
07
Draw the Parabola
With the vertex \((1, 0)\), axis of symmetry \(x = 1\), y-intercept \((0, -1)\), and additional points \((2, -1)\) and \((-1, -4)\), plot these on a coordinate plane. Because the parabola opens downwards, draw a smooth curve through these points to complete the graph of the equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parabola
A parabola is a type of curve on the graph that represents a quadratic function. It has a U-shape or an inverted U-shape depending on the sign of the leading coefficient in the quadratic equation. When dealing with a quadratic function in standard form, like \[ y = ax^2 + bx + c \], the shape of the parabola is primarily influenced by the coefficient, \(a\).
The parabola's symmetry and certain points that we calculate, like the vertex and y-intercept, guide us in accurately sketching it on a graph.
- If \(a > 0\), the parabola opens upwards forming a U-shape.
- If \(a < 0\), it opens downwards forming an inverted U-shape.
The parabola's symmetry and certain points that we calculate, like the vertex and y-intercept, guide us in accurately sketching it on a graph.
Vertex of a Quadratic Function
The vertex of a quadratic function is a key feature of a parabola. It represents the highest or lowest point on the graph, contingent on whether the parabola opens upward or downward. For the equation \[ y = ax^2 + bx + c \], the vertex can be found using the formula \[ h = -\frac{b}{2a} \]. Substituting the appropriate values from the equation \( y = -x^2 + 2x - 1 \), calculates:
\[ h = -\frac{2}{2(-1)} = 1 \]. By then substituting \( x = 1 \) back into the original quadratic equation, we find that \[ y = -(1)^2 + 2(1) - 1 = 0 \], making the vertex \((1, 0)\).
The vertex offers valuable insight. In a parabola that opens downward, such as in this exercise, the vertex is the maximum point. Understanding the vertex’s role is instrumental in graphing since it is a fixed point around which the parabola opens or closes.
\[ h = -\frac{2}{2(-1)} = 1 \]. By then substituting \( x = 1 \) back into the original quadratic equation, we find that \[ y = -(1)^2 + 2(1) - 1 = 0 \], making the vertex \((1, 0)\).
The vertex offers valuable insight. In a parabola that opens downward, such as in this exercise, the vertex is the maximum point. Understanding the vertex’s role is instrumental in graphing since it is a fixed point around which the parabola opens or closes.
Axis of Symmetry
The axis of symmetry is an invisible line that divides the parabola into two mirror-image halves. For a quadratic function, this line runs vertically through the vertex, reflecting the parabola's symmetry. When we have derived the vertex \((h, k)\), the x-coordinate provides the formula for the axis of symmetry's equation.
In the example \( y = -x^2 + 2x - 1 \), we found the vertex to be at \((1, 0)\). Consequently, the axis of symmetry can be written as \[ x = 1 \].
The axis of symmetry is super helpful because:
In the example \( y = -x^2 + 2x - 1 \), we found the vertex to be at \((1, 0)\). Consequently, the axis of symmetry can be written as \[ x = 1 \].
The axis of symmetry is super helpful because:
- It helps in graphing additional points easily.
- It ensures that the properties of symmetry are utilized, allowing one to draw an accurate parabolic graph.
- Every point on one side of the axis has a corresponding point on the opposite side.
Y-Intercept
The y-intercept is where a graph intersects the y-axis. For any function, this occurs where the value of \(x\) is zero. In terms of the parabola, it's a key point that aids in sketching its path on the graph.
For the quadratic equation \[ y = -x^2 + 2x - 1 \], setting \( x = 0 \) reveals:
\[ y = -(0)^2 + 2(0) - 1 = -1 \]. Hence, the y-intercept is point \((0, -1)\).
The y-intercept is quite useful:
For the quadratic equation \[ y = -x^2 + 2x - 1 \], setting \( x = 0 \) reveals:
\[ y = -(0)^2 + 2(0) - 1 = -1 \]. Hence, the y-intercept is point \((0, -1)\).
The y-intercept is quite useful:
- It offers a starting reference point when sketching the graph.
- Since the parabola is symmetric, it coordinates with the axis of symmetry for plotting additional points on the graph.