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Chapter 3: Problem 24

Use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range. $$f(x)=\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}$$

Short Answer

Expert verified
Vertex: (\(\frac{1}{2}, \frac{5}{4}\)), Intercepts: x-intercepts at \(x = 0, 1\) and y-intercept at \(f(0) = 0\), Axis of Symmetry: \(x = \frac{1}{2}\), Domain: \(-\infty < x < \infty\), Range: \(-\infty < y < \frac{5}{4}\).

Step by step solution

01

Identify the Vertex

The equation provided is already in vertex form which is \(f(x) = a(x - h)^2 + k\), where (h, k) is the vertex of the parabola. Comparing this to the given equation, \(f(x)=\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}\), we find that the vertex is \((\frac{1}{2}, \frac{5}{4})\).
02

Identify the Intercepts

To find the x-intercepts, we set \(f(x) = 0\), and solve for \(x\). To find the y-intercept, we set \(x = 0\), and solve for \(f(x)\). Solving these we find out that the x-intercepts are \(x = 0, 1\) and the y-intercept is \(f(0) = 0\).
03

Identify the Axis of Symmetry

The axis of symmetry for a parabola in the vertex form is the line \(x = h\). For our function, this is the line \(x = \frac{1}{2}\).
04

Sketch the Graph

Now we can sketch the graph of our function using the vertex, intercepts, and the axis of symmetry. The graph is a downward-opening parabola (since the leading coefficient of our quadratic term is negative) with vertex at (\(\frac{1}{2}, \frac{5}{4}\)), crossing the x-axis at \(x = 0, 1\), and symmetric about the line \(x = \frac{1}{2}\).
05

Identify the Domain and Range

The domain of a quadratic function is all real numbers, so the domain is \(-\infty < x < \infty\). Because the graph is a downward-opening parabola, the maximum point is the y-coordinate of the vertex. Hence, the range of the function is \(-\infty < y < \frac{5}{4}\).

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