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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. $$f(x)=2 x^{2}+4 x-3$$

Short Answer

Expert verified
The axis of symmetry is \(x=-1\). The x-intercepts are at \(x=-1-\sqrt{\frac{5}{2}}\) and \(x=-1+\sqrt{\frac{5}{2}}\). The domain of \(f(x)\) is \(\mathbb{R}\) and the range of \(f(x)\) is \([-5, +\infty)\).

Step by step solution

01

Write the Quadratic Function in Vertex Form

The standard form of a quadratic function is \(y=a(x-h)^2+k\), where \((h, k)\) is the vertex of the parabola. The given function is \(f(x)=2x^2+4x-3\). Using the method of completing the square, \(f(x)\) can be written in vertex form. \[f(x)=2(x^2+2x)-3=2(x^2+2x+1-1)-3=2[(x+1)^2-1]-3=2(x+1)^2-5\]. So, the vertex of the parabola is at (-1,-5) which is \((h,k)\).
02

Finding the x-intercepts

The x-intercepts are determined by solving for \(x\) when \(f(x)=0\). Doing so gives: \[0=2(x+1)^2-5 \Rightarrow (x+1)^2=\frac{5}{2}\] The solutions are \(x=-1-\sqrt{\frac{5}{2}}\) and \(x=-1+\sqrt{\frac{5}{2}}\). Therefore, the x-intercepts are \(-1-\sqrt{\frac{5}{2}}\) and \(-1+\sqrt{\frac{5}{2}}\)
03

Finding the Axis of Symmetry

The axis of symmetry is the vertical line \(x=h\), where \(h\) is the x-coordinate of the vertex. In this case, the axis of symmetry is \(x=-1\)
04

Identifying the Domain and the Range of the Function

The domain of any quadratic function is all real numbers, so the domain of \(f(x)\) is \(\mathbb{R}\). Since the coefficient of \(x^2\) is positive, the parabola opens upwards, which means the lowest point is the vertex. Therefore, the range is \([-5, +\infty)\)

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