Question

find the range of the quadratic function.
$y=2x^{2}+8x+5$
write your answer as an inequality using $x$ or $y$ as appropriate.
or, you may instead click on \empty set\ or \all reals\ as the answer.

Explanation:

Step1: Rewrite in vertex form

Complete the square for $y=2x^2+8x+5$.
First factor out 2 from the first two terms:
$y=2(x^2+4x)+5$
Add and subtract $(\frac{4}{2})^2=4$ inside the parentheses:
$y=2(x^2+4x+4-4)+5=2((x+2)^2-4)+5$
Simplify:
$y=2(x+2)^2-8+5=2(x+2)^2-3$

Step2: Analyze the vertex form

For $y=a(x-h)^2+k$, $a=2>0$, so the parabola opens upward, and the vertex $(h,k)=(-2,-3)$ is the minimum point.
The minimum value of $y$ is $-3$, and $y$ can increase to infinity.

Answer:

$y\geq -3$