Question

How do you find the vertex and intercepts for `y=x^2 + 4x + 2`?

Answer 1

Complete the square. `y=(x+4/2)^2-2^2+2=(x+2)^2-2`. So the vertex is at `(-2,-2)`

Explanation:

The minimum possible value of `(x+2)^2` is zero and will occur when `x=-2`. To complete the square, add half the coefficient of `x` to the `x^2` term, square the sum, and take away the square of the number you added. thus returning the expression to its original value.

The `y`-intercept is `(0,2)`. The `x`-co-ordinates of the `x`-intercepts are the roots of the equation `x^2+4x+2=0`, namely `-2+-sqrt(2)`.