Question

27/32 identify the form of the quadratic function: f(x)=2x^2 + 4x + 1 vertex standard factored quadratic

Explanation:

Step1: Recall quadratic - form definitions

The given function \(f(x)=2x^{2}+4x + 1\) is already in standard form \(y = ax^{2}+bx + c\), where \(a = 2\), \(b = 4\), and \(c = 1\).

Step2: Convert to vertex - form

Complete the square. First, factor out the coefficient of \(x^{2}\) from the first two terms: \(f(x)=2(x^{2}+2x)+1\). Then, complete the square inside the parentheses. For \(x^{2}+2x\), we add and subtract \((\frac{2}{2})^{2}=1\) inside the parentheses: \(f(x)=2(x^{2}+2x + 1-1)+1=2((x + 1)^{2}-1)+1=2(x + 1)^{2}-2 + 1=2(x + 1)^{2}-1\). The vertex - form is \(y=a(x - h)^{2}+k\), and here \(h=-1\), \(k=-1\), \(a = 2\).

Step3: Check factored form

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) to find the roots. For \(a = 2\), \(b = 4\), \(c = 1\), \(x=\frac{-4\pm\sqrt{4^{2}-4\times2\times1}}{2\times2}=\frac{-4\pm\sqrt{16 - 8}}{4}=\frac{-4\pm\sqrt{8}}{4}=\frac{-4\pm2\sqrt{2}}{4}=\frac{-2\pm\sqrt{2}}{2}\). So the factored form is \(f(x)=2(x-\frac{-2+\sqrt{2}}{2})(x-\frac{-2 - \sqrt{2}}{2})\).

Answer:

Standard form: \(f(x)=2x^{2}+4x + 1\)
Vertex form: \(f(x)=2(x + 1)^{2}-1\)
Factored form: \(f(x)=2(x-\frac{-2+\sqrt{2}}{2})(x-\frac{-2 - \sqrt{2}}{2})\)