How do you graph the parabola # y = x^2 + 4x + 1# using vertex, intercepts and additional points?

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Answer 1 · 2018-07-20T05:21:06

Vertex #(-2, -3)#
Y-intercept #(0,1)#
x-intercepts #(0.268, 0); (-3.732, 0)#

Given -

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

Vertex

#x=(-b)/(2a)=(-4)/2=-2#

At #x=-2; y=(-2)^2+4(-2)+1=4-8+1=-3#

Vertex #(-2, -3)#

Y-intercept
At #x=0; y=(0)^2+4(0)+1=0+0+1=1#

Y-intercept #(0,1)#

x_intercept
At #y=0; x^2+4x-1=0#

#x^2+4x=-1#

#x^2+4x+4=-1+4=3#

#(x+2)^2=3#

#x+2=+-sqrt3=+-1.732#

#x=1.732-2=0.268#

#x=-1.732-2=-3.732#

x-intercepts #(0.268, 0); (-3.732, 0)#

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