Given equation: y =2x^2+12x+13.
We need to find the axis of symmetry and the coordinates of the vertex of the graph of the function.
The formula of axis of symmetry is : [tex]-\frac{b}{2a}[/tex].
a= 2 and b=12.
Therefore, [tex]-\frac{b}{2a} = -\frac{12}{2(2)} = -\frac{12}{4} = -3[/tex].
Let us find y-coordinate of the vertex.
Plugging x=-3 in given quadratic y =2x^2+12x+13.
y= 2(-3)^2+12(-3)+13 = 2(9) -36 +13 = 18-36 +13 = -5.
We got x-coordinate of the vertex -3 and y-coordinate of the vertex -5.
