How do you find the axis of symmetry, vertex and x intercepts for #y=x^2+4x+6#?

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Answer 1 · 2018-07-27T15:04:37

#x+2=0, \ \ (-2, 2)# & no x-intercept

The given equation:

#y=x^2+4x+6#

#y=(x^2+4x+4)+2#

#y=(x+2)^2+2#

#(x+2)^2=y-2#

Above equation is in vertex form of upward parabola:

#(x-x_1)^2=4a(y-y_1)# which has

Axis of symmetry: #x-x_1=0#

#\implies x+2=0#

Vertex: #(x_1, y_1)\equiv(-2, 2)#

The parabola will intersect the x-axis where #y=0#

#\therefore x^2+4x+6=0#

#b^2-4ac=4^2-4(1)(6)=-8<0#

Above quadratic equation has no real roots i.e. the parabola doesn't intersect the x-axis.

hence no x-intercept