What is the range of y = 3x^2 - 7?

Aug 16, 2016

Since your function is a quadratic and the quadratic coefficient is positively-signed, it has a single minimum value, and extends upwards towards $\infty$.

Therefore, when you've found the smallest possible y value, that's your lower bound.

Since the second degree term is not horizontally shifted (that is, ${\left(x + c\right)}^{2}$, where $c = 0$), the minimum is when $x = 0$.

$\textcolor{b l u e}{\left[- 7 \text{, } \infty\right)}$

Another way is to see that

$f \left(x\right) + 7 = 3 {x}^{2} \ge 0$

$f \left(x\right) + 7 \ge 0$

$f \left(x\right) \ge - 7$

Hence the range is ${R}_{f} = \left[- 7 , + \infty\right)$