Given the function:
[tex]f(x)=2x^2+4x+1[/tex]
The coefficients of the quadratic equations are:
[tex]\begin{gathered} A=2 \\ B=4 \\ C=1 \end{gathered}[/tex]
Where C is the y-intercept, which in this case is 1.
To find the vertex, we can write the vertex form of the equation. Completing squares:
[tex]\begin{gathered} f(x)=2x^2+4x+1=2(x^2+2x)+1 \\ \Rightarrow2(x^2+2x)+1=2(x^2+2x+1-1)+1=2((x+1)^2-1)+1 \\ \Rightarrow f(x)=2(x+1)^2-1 \end{gathered}[/tex]
From the generic vertex form of a quadratic equation:
[tex]f(x)=a(x-h)^2+k[/tex]
We can identify:
Vertex → (h, k)
If a > 0: The parabola opens up
If a < 0: The parabola opens down
x = h is the equation of the axis of symmetry
From our problem:
[tex]\begin{gathered} a=2 \\ h=-1 \\ k=-1 \end{gathered}[/tex]
So this parabola opens up, its vertex is at (-1, -1), and the axis of symmetry is
x = -1.
The strategic points are the vertex, the y-intercept, and those values that make f(x) = 0.
[tex]\begin{gathered} 2(x+1)^2-1=0 \\ (x+1)^2=\frac{1}{2} \\ x+1=\pm\frac{1}{\sqrt[]{2}} \\ x=\pm\frac{1}{\sqrt[]{2}}-1 \end{gathered}[/tex]
Then, the two additional strategic points are:
[tex]\begin{gathered} (\frac{1}{\sqrt[]{2}}-1,0) \\ (-\frac{1}{\sqrt[]{2}}-1,0) \end{gathered}[/tex]
