How do you graph # y = x^2 + 4x + 1#?

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Answer 1 · 2018-06-01T14:49:39

Find the intercepts and the vertex.

Set x=0 to find the y intercept:

#y = x^2 + 4x + 1#

#y = 0^2 + 0x + 1#

#y=1#

Set y=0 to find the x intercept(s) if they exist:

#y = x^2 + 4x + 1#

#0 = x^2 + 4x + 1#

#x = (-b+-sqrt(b^2 -4ac))/(2a)#

#x = (-4+-sqrt(4^2 -4*1*1))/(2*1)#

#x=2+-sqrt3#

Now find the Axis of Symmetry (aos):

#aos = -b/(2a) = (-4)/(2*1) = -2#

Find the vertex:

vertex = (aos, f(aos)) #= (-2, f(-2))#

#f(x) = x^2 + 4x + 1#

#f(-2) = (-2)^2 + 4* -2 + 1 = -3#

vertex #= (-2, -3)#

Finally since a > 0 the parabola open up like a smile so the vertex is a minimum, now graph:

graph{x^2 +4x+1 [-11, 9, -3.64, 6.36]}