How do you find the maximum/ minimum of the quadratic equation: f(x)=x^2+2x+4

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Question:

How do you find the maximum/ minimum of the quadratic equation: f(x)=x^2+2x+4

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if you only know algebar, find the vertexwe see it opens upso minimumso the x value of the vertex is -b/2a for ax^2+bx+c=ysof(x)=1x^2+2x+4x value of vertex is -2/(2*1)=-2/2=-1x value of the minimum is x=-1if we find f(-1) we get 3minimum value is 3 at x=-1

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