Question

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

Answer 1

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)`