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Chapter 3: Problem 43

Identify the vertex, axis of symmetry, y-intercept, x-intercepts, and opening of each parabola, then sketch the graph. $$y=x^{2}-x$$

Short Answer

Expert verified
Vertex: (0.5, -0.25), Axis of Symmetry: x = 0.5, Y-intercept: (0, 0), X-intercepts: (0, 0) and (1, 0), Opens upwards.

Step by step solution

01

- Identify the vertex

To find the vertex of the parabola given by the equation \(y = x^2 - x \), rewrite it in the vertex form \(y = a(x-h)^2 + k\). Start by completing the square.\[y = x^2 - x + \frac{1}{4} - \frac{1}{4}\]This simplifies to:\[y = (x - \frac{1}{2})^2 - \frac{1}{4}\]So, the vertex form is \(y = (x - 0.5)^2 - 0.25\) and the vertex is at \((0.5, -0.25)\).
02

- Find the axis of symmetry

The axis of symmetry is a vertical line that runs through the vertex of the parabola. Since the vertex is at \((0.5, -0.25)\), the axis of symmetry is \(x = 0.5\).
03

- Determine the y-intercept

To find the y-intercept, set \(x = 0\) in the original equation: \[y = 0^2 - 0 = 0\]So, the y-intercept is at \((0, 0)\).
04

- Calculate the x-intercepts

To find the x-intercepts, set \(y = 0\) in the equation and solve for \(x\):\[0 = x^2 - x\]Factor the right side:\[0 = x(x - 1)\]Setting each factor to zero gives \(x = 0\) or \(x = 1\). So, the x-intercepts are at \((0, 0)\) and \((1, 0)\).
05

- Determine the opening

The sign of the coefficient of the \(x^2\) term (which is positive 1) tells the direction of the parabola's opening. Since it is positive, the parabola opens upwards.
06

- Sketch the graph

Draw the graph using the identified key features: vertex \((0.5, -0.25)\), axis of symmetry \(x = 0.5\), y-intercept \((0, 0)\), and x-intercepts \((0, 0)\) and \((1, 0)\). Sketch a parabola opening upwards.

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Key Concepts

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

axis of symmetry
The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. For any parabola, this line runs through the vertex. In the equation of the form \(y = ax^2 + bx + c\), the x-coordinate of the vertex gives us the axis of symmetry.

For our equation \(y = x^2 - x\), we first find the vertex, which is at \((0.5, -0.25)\). Once we have the vertex, the x-coordinate of the vertex is the equation for the axis of symmetry. Hence, the axis of symmetry for this parabola is \(x = 0.5\).

The axis of symmetry is crucial because it helps in sketching the graph accurately. It allows us to draw equal distances from the vertex on either side along the x-axis, ensuring our parabola is symmetrical.
y-intercept
The y-intercept is the point where the parabola crosses the y-axis. This happens when \(x = 0\). To find the y-intercept, simply substitute \(x = 0\) in the given equation.

For the equation \(y = x^2 - x\):
\(y = 0^2 - 0 = 0\)

Hence, the y-intercept is at \((0, 0)\).

The y-intercept serves as a starting point for sketching the graph and provides one of the critical points through which the parabola passes. It is typically one of the simpler points to calculate but is essential for plotting the graph accurately.
x-intercepts
The x-intercepts are the points where the parabola crosses the x-axis. This occurs when \(y = 0\). To find them, set the equation equal to zero and solve for \(x\).

For the equation \(y = x^2 - x\):

\(0 = x^2 - x\)
Factorize the right side:
\(0 = x(x - 1)\)
Setting each factor equal to zero, we get:

\(x = 0 \text{ or } x = 1\)

Therefore, the x-intercepts are at \((0, 0)\) and \((1, 0)\).

These points are vital when sketching the graph as they help in identifying points through which the parabola passes and understanding its shape.
parabola opening direction
The direction in which a parabola opens (upward or downward) depends on the coefficient of the \(x^2\) term. If the coefficient (a) is positive, the parabola opens upwards. If the coefficient is negative, the parabola opens downwards.

For the equation \(y = x^2 - x\), the coefficient of \(x^2\) is 1, which is positive. Therefore, the parabola opens upwards.

Knowing the direction of the parabola is essential for sketching it correctly. An upward opening means the vertex represents the minimum point on the graph, and the arms of the parabola extend infinitely upwards.

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