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Chapter 3: Problem 31

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. $$f(x)=2 x-x^{2}+3$$

Short Answer

Expert verified
The vertex of the function is (1, 4), with a y-intercept at (0, 3). There are no x-intercepts. The graph of the function is a downwards-opening parabola. The equation of the axis of symmetry is \(x = 1\). The domain of the function is \(-\infty, +\infty\) and the range is \(-\infty, 4\].

Step by step solution

01

Identify Vertex

The vertex of a parabola \(y = a(x - h)^2 + k\) is \((h, k)\). In standard form \(y = ax^2 + bx + c\), the x-coordinate of the vertex can be found using the formula \(h = -b/2a\). Our function is \(f(x) = 2x - x^2 + 3\). We rewrite it in the form \(f(x) = -(x^2 - 2x) + 3\), so a = -1, b = 2. Thus, the x-coordinate of the vertex is \(h = -b/2a = -2/2*(-1) = 1\). We substitute \(x = 1\) into the function to find the y-coordinate: \(f(1) = 2*1 - 1^2 + 3 = 4\). Therefore, the vertex is (1, 4).
02

Identify Intercepts

To find the x-intercepts, we set \(f(x) = 0\), obtaining the equation \(0 = 2x - x^2 + 3\). This simplifies to \(x^2 - 2x + 3 = 0\). Solving it using the quadratic formula does not yield real roots meaning there are no x-intercepts. The y-intercept is found by evaluating the function at \(x = 0\): \(f(0) = 2*0 - 0 + 3 = 3\). Therefore, the y-intercept is (0, 3).
03

Draw the Graph

We plot the vertex (1, 4), and the y-intercept (0, 3), noting there are no x-intercepts. The parabola opens downward because the coefficient of \(x^2\) is negative. We sketch the parabola accordingly.
04

Axis of Symmetry

The axis of symmetry of a parabola is a vertical line through the vertex, given by \(x = h\). From step 1, we know the vertex is (1, 4), hence the equation of the axis of symmetry is \(x = 1\).
05

Determine Domain and Range

By nature, the domain of any quadratic function is all real numbers, so the domain is \(-\infty, +\infty\). The range of the function is all the possible output values (y-values). As our parabola opens downwards with a maximum point at (1, 4), the range is \(-\infty, 4\].

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