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Chapter 3: Problem 32

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. $$f(x)=5-4 x-x^{2}$$

Short Answer

Expert verified
The vertex of the parabola is at (-2, 9). Its x-intercepts are at x=-1 and x=5, and the y-intercept is at y=5. The axis of symmetry is x=-2. The function's domain is all real numbers, and its range is \(-鈭� < y 鈮� 9\).

Step by step solution

01

Identify the vertex

The vertex form of a parabola is \(f(x)=a(x-h)^2+k\), where \((h,k)\) is the vertex of the parabola. Shifting the given function to the vertex form by completing the square results in \(f(x)=-1*(x+2)^{2}+9\). Therefore, the vertex is \((-2,9)\).
02

Calculate intercepts and axis of symmetry

The x-intercepts can be found by setting \(f(x)=0\). Solving for x in the equation \(0=5-4x-x^2\) gives \(x=-1\) and \(x=5\). The y-intercept is found by evaluating \(f(x)\) at \(x=0\), which gives \(y=f(0)=5\). The equation for the axis of symmetry in a parabola is \(x=h\), with \(h\) being the x-coordinate of the vertex. Thus the axis of symmetry here is \(x=-2\).
03

Determine domain and range

The domain of a function is the set of all possible input values (x-values), which for a quadratic function is all real numbers, or \(-鈭� < x < 鈭瀄). The range of a function is the set of all possible output values (y-values). Since the parabola opens downward (the coefficient of the \(x^2\) term is negative), the maximum point is the vertex. Thus, the range is \(-鈭� < y 鈮� 9\).

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