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Graph. $$y=-x^{2}+2 x-1$$

Short Answer

Expert verified
The vertex is (1, 0), the axis of symmetry is x = 1, and the y-intercept is (0, -1). The parabola opens downwards.

Step by step solution

01

Identify the key components

The given quadratic equation is in the form of a standard quadratic function: \( y = -x^2 + 2x - 1 \). Notice that the leading coefficient is negative, which indicates that the parabola opens downwards.
02

Find the vertex

To find the vertex of the parabola, use the vertex formula for a quadratic function, \( x = - \frac{b}{2a} \). In this equation, \( a = -1 \), \( b = 2 \). Hence, \( x = - \frac{2}{2(-1)} = 1 \). Substitute \( x = 1 \) back into the equation to find \( y \): \( y = -(1)^2 + 2(1) - 1 \), which simplifies to 0. So, the vertex is \( (1,0) \).
03

Determine the axis of symmetry

The axis of symmetry for the parabola can be found using the x-coordinate of the vertex, which is \( x = 1 \). Therefore, the axis of symmetry is the vertical line \( x = 1 \).
04

Find the y-intercept

To find the y-intercept, set \( x = 0 \) in the equation: \( y = -(0)^2 + 2(0) - 1 \). Simplified, this gives \( y = -1 \). So, the y-intercept is \( (0, -1) \).
05

Find additional points

To sketch the graph accurately, calculate a few more points around the vertex. For example, choose \( x = -1 \) and \( x = 2 \): \( y = -(-1)^2 + 2(-1) - 1 = -1 - 2 - 1 = -4 \) gives the point \( (-1, -4) \); and \( y = -(2)^2 + 2(2) - 1 = -4 + 4 - 1 = -1 \) gives the point \( (2, -1) \).
06

Draw the graph

Plot the points found: \( (1, 0) \), \( (0, -1) \), \( (-1, -4) \), and \( (2, -1) \). Draw a smooth curve to form the parabola passing through these points and ensure it opens downwards. The vertex \( (1, 0) \) should be at the highest point of the parabola.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

vertex of a quadratic
The vertex of a quadratic function is the highest or lowest point on its graph, known as a parabola. In general, for a quadratic function in the form \( y = ax^2 + bx + c \), the vertex can be found using the formula \( x = -\frac{b}{2a} \).

Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate.

For the equation \( y = -x^2 + 2x - 1 \), we plug in the values from the equation: \( a = -1 \) and \( b = 2 \).

This gives us \( x = -\frac{2}{2(-1)} = 1 \). By substituting \( x = 1 \) back into the equation, we find that \( y = -(1)^2 + 2(1) - 1 = 0 \).

Hence, the vertex of the parabola is \( (1, 0) \). The vertex is a critical point because it helps in determining the maximum or minimum value of the function and overall shape of the graph.
axis of symmetry
The axis of symmetry of a quadratic function is a vertical line that passes through the vertex, dividing the parabola into two mirror-image halves. For the standard quadratic function \( y = ax^2 + bx + c \), the axis of symmetry can be determined from the x-coordinate of the vertex.

In the given equation \( y = -x^2 + 2x - 1 \), we already calculated the vertex to be \( (1, 0) \).

Therefore, the axis of symmetry is the vertical line \( x = 1 \).

The axis of symmetry is significant because it provides a line of reflection for the parabola, ensuring that for every point on the left side of the line, there is a corresponding point on the right at an equal distance. It helps in graphing the quadratic because once one half of the parabola is plotted, the other half can be easily mirrored.
y-intercept
The y-intercept is the point where the graph of the function crosses the y-axis. For quadratic functions in the form \( y = ax^2 + bx + c \), to find the y-intercept, simply set \( x = 0 \) and solve for \( y \).

In the given equation \( y = -x^2 + 2x - 1 \), substituting \( x = 0 \) yields \( y = -(0)^2 + 2(0) - 1 = -1 \). Hence, the y-intercept is \( (0, -1) \).

The y-intercept is crucial for graphing as it provides an initial starting point for the graph. Knowing the y-intercept helps in sketching the parabola accurately and gives insight into where the parabola touches the y-axis.
parabola
A parabola is the graph of a quadratic function and can take on different shapes depending on the coefficients of the equation. In general, for quadratic functions of the form \( y = ax^2 + bx + c \), the parabola can either open upwards or downwards.

If \( a \) is positive, the parabola opens upwards. If \( a \) is negative, as in the equation \( y = -x^2 + 2x - 1 \), the parabola opens downwards.

The shape and direction of the parabola also depend on the vertex, which we previously found to be \( (1, 0) \), and the axis of symmetry, which is \( x = 1 \).

Additionally, key points such as the y-intercept \( (0, -1) \) and any other points calculated (like \( (-1, -4) \) and \( (2, -1) \)) help in drawing the overall curve.

Understanding the properties of a parabola, such as its vertex, axis of symmetry, and intercepts, is essential in graphing quadratic functions accurately and appreciating their geometric beauty.

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