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

1 Answer
Roy E.
Dec 23, 2016

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)#.

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