Question

find the range of the quadratic function.
y = 3x² - 12x + 13
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

$$\begin{align*} y&=3x^2 - 12x + 13\\ &=3(x^2 - 4x) + 13\\ &=3(x^2 - 4x + 4 - 4) + 13\\ &=3[(x-2)^2 - 4] + 13\\ &=3(x-2)^2 - 12 + 13\\ &=3(x-2)^2 + 1 \end{align*}$$

Step2: Analyze squared term range

For all real $x$, $(x-2)^2 \geq 0$

Step3: Find $y$ range

Multiply by 3: $3(x-2)^2 \geq 0$
Add 1: $3(x-2)^2 + 1 \geq 1$, so $y \geq 1$

Answer:

$y \geq 1$