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Chapter 3: Problem 38

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. $$f(x)=6-4 x+x^{2}$$

Short Answer

Expert verified
The vertex of the parabola is (2,2), the parabola's axis of symmetry is \(x = 2\), the domain of the function is all real numbers and the range of the function is \(y 鈮� 2\).

Step by step solution

01

Finding the Vertex

The vertex of a quadratic equation in the standard form \(f(x) = a(x-h)^2 + k\) is given by the point (h,k). Here, our equation is in the form \(f(x) = x^2 - 4x + 6\). We complete the square for the equation to find h and k. The h value is found using the formula \(-b/2a\), where a and b are coefficients of \(x^2\) and \(x\) respectively. In this case, a=1 and b=-4, hence the h value is calculated as: \(h = -(-4) / 2*1 = 2\). We substitute \(h = 2\) in \(f(x)\) to find k. So, the vertex of the parabola is (2,2)
02

Finding the Intercepts and Axis of Symmetry

The intercepts are found by setting \(y = 0\) in the formula and solving for \(x\). These calculations yield two x-values which represent the x-intercepts of the graph. For the function \(f(x) = x^2 - 4x + 6\), setting \(f(x) = 0\) gives: \(0 = x^2 - 4x + 6\), which can be solved using the quadratic formula. From the vertex (h,k), we learn the axis of symmetry of a parabola is always \(x = h\) so here \(x = 2\).
03

Determine the Domain and Range

The domain of a function is the set of all possible x-values and for any quadratic function, the domain is all real numbers so \(x 鈭� R\). The range of a function, on the other hand, is the set of possible y-values. Since our equation opens upwards and the vertex form tells us that the y-coordinate of the vertex is the minimum value of \(f(x)\), thus the range is \(y 鈮� 2\).

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