Identify the vertex of the parabola defined by \(\displaystyle \L \;y\,=\,2x^2\,+\,8x\,+\,9\).
Given a quadratic in the form: \(\displaystyle ax^2\,+\,bx\,+\,c\)
The vertex is: \(\displaystyle \L \;(\frac{\,-\,b}{2a}\,,\,f(\frac{\,-\,b}{2a}))\)
So we find b and a to get the x coordinate of the vertex:
\(\displaystyle \L \;\frac{\,-\,b}{2a}\,=\,\frac{\,-\,8}{2(2)}\,=\,-\,2\)
The we plug that back in to get the y coordinate of the vertex.
So solve: \(\displaystyle y\,=\,2(\,-\,2)^2\,+\,8(\,-\,2)\,+\,9\)
To get an answer \(\displaystyle (\,-\,2\,,\,y)\)
For the second, find the x coordinate of the vertex and put x = that and you'll have the line of symmetry.