Question

find the range of the quadratic function.
$y = 2x^2 + 12x + 17$
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: Complete the square

$$\begin{align*} y&=2x^2+12x+17\\ &=2(x^2+6x)+17\\ &=2(x^2+6x+9-9)+17\\ &=2((x+3)^2-9)+17 \end{align*}$$

Step2: Simplify the expression

$$\begin{align*} y&=2(x+3)^2-18+17\\ &=2(x+3)^2-1 \end{align*}$$

Step3: Analyze the squared term

Since $(x+3)^2\geq0$ for all real $x$, multiply by 2: $2(x+3)^2\geq0$

Step4: Find the minimum value of $y$

Subtract 1: $2(x+3)^2-1\geq -1$, so $y\geq -1$

Answer:

$y\geq -1$