Question

Determine the vertex and the axis of symmetry for the function below: y=x^2+4x+1

Answer 1

`\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\ \begin{array}{llccll} y = &{{ 1}}x^2&{{ +4}}x&{{ +1}}\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)`

the squared variable is the "x", thus is a vertical parabola, thus, the axis of symmetry will occur over the x-coordinate of the vertex, and thus will be
`\bf x=-\cfrac{{{ b}}}{2{{ a}}}`