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

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