How do you graph #y<=x^2+8x+16#?

1 Answer
Jun 6, 2017

Answer:

It is all the ordered pairs on the line of and outside of the plot of #y=x^2+8x+16#. The line is solid as we have #>=#

Explanation:

#color(blue)("Determine the key features of "y=x^2+8x+16)#

As the #x^2# term is positive the graph is of form #uu# thus the vertex is a minimum

Note that #4^2=16 and 4+4=8#

so #y=0=(x+4)(x+4)# This is called duality

Thus the graph is such that the x-axis is tangential thus
#y_("vertex")=0#

Consider the #x# term: we have #+8x#

Compare to the equation standardised form of #y=ax^2+bx+c#
From this #x_("vertex")=(-1/2)xxb/a#

Thus #x_("vertex")=(-1/2)xx8/1=-4" "# as in keeping with the above.

Vertex#->(x,y)=(-4,0)#

y-intercept is the constant 16#->(x,y)=(0,16)#

Tony B