Given
The function,
[tex]y=x^2+8x+16[/tex]To find:
The range.
Explanation:
It is given that,
[tex]y=x^2+8x+16[/tex]That implies,
Since, For a parabola, ax²+bx+c, with vertex (h,k).
Then,
[tex]\mathrm{If}\:a>0\:,\mathrm{the\:range\:is}\:f\left(x\right)\ge\:k[/tex]Therefore,
[tex]\begin{gathered} y=x^2+8x+16 \\ \Rightarrow y=(x+4)^2-16+16 \\ \Rightarrow y=(x+4)^2 \end{gathered}[/tex]Hence, the vertex is (-4,0).
And, a=1.
Then,
[tex]y\ge0[/tex]Hence, the range is [0,∞).
