How do you graph #f(x)=(x+4)^2# and identify the x intercepts, vertex?

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Answer 1 · 2018-02-03T03:31:49

#f(x)# has the graph of standard parabolic #y=x^2# shifted 4 units negative on the #x-#axis.

#x-#intercept and vertex at #(-4,0)#

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

Let #x' = x+4 -> x=x'-4#

#f(x) = (x')^2#

Hence, #f(x)# has the graph of standard #y=x^2# shifted 4 units negative ('left') on the #x-#axis

The #x-#intercepts occur where #f(x)=0#

I.e. where #(x+4)^2 = 0#

#(x+4)(x+4)=0#

#->f(x)# has coincident #x-#intercepts at #(-4,0)#

Expanding #f(x)#

#f(x)=x^2+8x+16#

#f(x)# is a parabola of the form #ax^2+bx+c# with vertex at #x=(-b)/(2a)#

Hence, the vertex of #f(x)# is at #x=(-8)/2 =-4#

So, the vertex of #f(x)# is also at (-4,0)

The graph of #f(x)# is shown below.

graph{(x+4)^2 [-10, 10, -5, 5]}