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

Ask by Lambert Ball. in the United States
Feb 25,2025

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The parabola \( y = x^{2} - x \) has the following features: - **Vertex**: \((\frac{1}{2}, -\frac{1}{4})\) - **Axis of Symmetry**: \( x = \frac{1}{2} \) - **y-intercept**: \((0, 0)\) - **x-intercepts**: \((0, 0)\) and \((1, 0)\) - **Opening**: Upward To sketch the graph, plot the vertex, draw the axis of symmetry, mark the intercepts, and plot additional points symmetrically to form a U-shaped curve opening upwards.

Solución

We’re given the quadratic function   y = x² – x. Below are the steps to analyze its features: 1. • Vertex:   – Write the quadratic in the form y = ax² + bx + c where a = 1, b = –1, and c = 0.   – The x-coordinate of the vertex is found by     h = –b/(2a) = –(–1)/(2·1) = 1/2.   – Substitute x = 1/2 back into the equation to find the y-coordinate:     y = (1/2)² – (1/2) = 1/4 – 1/2 = –1/4.   – Thus, the vertex is at (1/2, –1/4). 2. • Axis of Symmetry:   – The axis of symmetry is the vertical line that passes through the vertex.   – It is given by x = 1/2. 3. • y-intercept:   – The y-intercept occurs when x = 0.   – Substitute x = 0 into the equation: y = 0² – 0 = 0.   – So the y-intercept is at (0, 0). 4. • x-intercepts:   – The x-intercepts occur when y = 0.   – Set up the equation: 0 = x² – x.   – Factor: 0 = x(x – 1).   – Solve: x = 0 or x – 1 = 0 → x = 1.   – Therefore, the x-intercepts are at (0, 0) and (1, 0). 5. • Opening of the Parabola:   – Since the coefficient a = 1 is positive, the parabola opens upward. 6. • Sketching the Graph:   – Plot the vertex at (1/2, –1/4).   – Draw the axis of symmetry as a dashed vertical line through x = 1/2.   – Plot the intercepts: the y-intercept and one of the x-intercepts are both at (0, 0) while the other x-intercept is at (1, 0).   – Notice that (0, 0) is on the parabola and by symmetry the point (1, 0) is its mirror image about the line x = 1/2.   – For additional points, choose values equidistant from the axis. For example:     • When x = –0.5 (which is 1 unit left of x = 0.5),       y = (–0.5)² – (–0.5) = 0.25 + 0.5 = 0.75.     • Its symmetric point is x = 1.5 (1 unit right of x = 0.5), yielding:       y = (1.5)² – 1.5 = 2.25 – 1.5 = 0.75.   – Plot these points and sketch the smooth, curved U-shaped graph opening upward. Summary of key features:   • Vertex: (1/2, –1/4)   • Axis of symmetry: x = 1/2   • y-intercept: (0, 0)   • x-intercepts: (0, 0) and (1, 0)   • Opening: Upward This analysis provides all the necessary details to sketch an accurate graph of the parabola y = x² – x.

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Beyond the Answer

To find the vertex of the parabola given by the equation \( y = x^2 - x \), we can complete the square or use the vertex formula. The formula \( x = -\frac{b}{2a} \) gives us the x-coordinate of the vertex, where \( a = 1 \) and \( b = -1 \). So, \( x = -\frac{-1}{2 \cdot 1} = \frac{1}{2} \). Substituting \( x = \frac{1}{2} \) back into the equation, \( y = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = -\frac{1}{4} \). Thus, the vertex is \( \left(\frac{1}{2}, -\frac{1}{4}\right) \). The axis of symmetry is the vertical line that passes through the vertex, which is given by \( x = \frac{1}{2} \). To find the \( y \)-intercept, set \( x = 0 \): \( y = 0^2 - 0 = 0\), giving us the \( y \)-intercept at \( (0, 0) \). For the \( x \)-intercepts, set \( y = 0 \): \( 0 = x^2 - x \), which factors to \( x(x - 1) = 0 \). Thus, the \( x \)-intercepts are \( (0, 0) \) and \( (1, 0) \). The parabola opens upwards because the leading coefficient \( a = 1 \) is positive. Now, for a sketch: plot the vertex at \( \left(\frac{1}{2}, -\frac{1}{4}\right) \), draw the axis of symmetry as a dashed line through \( x = \frac{1}{2} \), mark the \( y \)-intercept at \( (0, 0) \) and the two \( x \)-intercepts at \( (0, 0) \) and \( (1, 0) \). The graph will be symmetric across the line \( x = \frac{1}{2} \), forming a classic upward-opening parabola. As a fun fact, parabolas have appeared in art and architecture for centuries! The ancient Greeks studied the properties of conic sections, and this shape can be seen in the design of certain amphitheaters and bridges, where the arch shape optimally distributes weight. For anyone diving deeper into quadratic functions, exploring the discriminant can be quite enlightening! The discriminant \( b^2 - 4ac \) helps you determine the number of real roots (intercepts) the quadratic has. If it’s positive, you get two real roots; if zero, one real root; and if negative, no real roots. It's like a mathematical crystal ball!

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