Question

What's the range of x^2+8x+16

Answer 1

Given

The function,

`y=x^2+8x+16`

To find:

The range.

Step-by-step explanation:

It is given that,

`y=x^2+8x+16`

That implies,

Since, For a parabola, ax²+bx+c, with vertex (h,k).

Then,

`\mathrm{If}\:a>0\:,\mathrm{the\:range\:is}\:f\left(x\right)\ge\:k`

Therefore,

`\begin{gathered} y=x^2+8x+16 \\ \Rightarrow y=(x+4)^2-16+16 \\ \Rightarrow y=(x+4)^2 \end{gathered}`

Hence, the vertex is (-4,0).

And, a=1.

Then,

`y\ge0`

Hence, the range is [0,∞).