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Chapter 3: Problem 37

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

Short Answer

Expert verified
The vertex of the quadratic function is (1, -1), the intercepts are (-1, 0), (2, 0) and (0, -2), the axis of symmetry is \(x = 1\), the domain is all real numbers \(-\infty < x < \infty\), and the range is \(-\infty < y 鈮� -1\).

Step by step solution

01

Finding the Vertex

The vertex form of a quadratic function is given by \(f(x) = a(x-h)^{2}+k\) where the vertex is \((h,k)\). But first, the equation should be rewritten in the form \(f(x) = -1(x^2-2x)+2\).The vertex of the quadratic function \(f(x)=2x - x^{2} - 2\) can be found by using the formula for h, which is \(h = -b/2a\). In this case, a=-1 and b=2. Hence, \(h = -2/(2*-1) = 1\). We substitute x = 1 into the equation to get the y-coordinate of the vertex as \(f(1) = 2*1 - (1)^{2} - 2 = -1\). So, the vertex is (1,-1).
02

Finding the Intercept

The x-intercepts of a quadratic function can be found by setting y (or f(x)) to zero and solving for x. For our example, we set \(f(x) = 2x - x^{2} - 2 = 0\). The roots of the equation thus obtained provide the x-intercepts. By solving, we get \(x = -1, 2\). The y-intercept is obtained by substituting \(x = 0\), which yields \(y =-2\). So, the intercepts are (-1 , 0), (2 , 0), and (0 , -2).
03

Finding the Axis of Symmetry

The axis of symmetry is the line that passes through the vertex of the parabola. It can be given by the equation \(x=h\). So, the axis of symmetry of the parabola is \(x = 1\).
04

Determining the Domain and Range

The domain of a function is the set of all possible input values (typically x-values), and since quadratic functions continue indefinitely in both directions, the domain is all real numbers, \(-\infty < x < \infty\). The range of a function is the set of all possible output values (typically y-values). Our function opens downwards because the coefficient of the x-squared term is negative. Hence, its maximum value is at the y-coordinate of the vertex, and it goes down to negative infinity from there. So, the range of our function is \(-\infty < y 鈮� -1\).
05

Sketching the Graph

Start by plotting the vertex,(1,-1) and the intercepts (-1, 0), (2, 0) and (0, -2). The vertex is the peak of the graph because it opens downwards. Understand that it is symmetric about axis \(x = 1\), so the graph to the left of \(x = 1\) is a mirror image of the graph to the right. You should also realise that the function decreases on either side of \(x = 1\). Now, join the vertex and intercepts with a smooth curve to complete the graph.

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