# How do you find the range of a quadratic equation f(x) = -x^2 + 14x - 48?

Apr 10, 2015

Given $f \left(x\right) = - {x}^{2} + 14 x - 48$

$f ' \left(x\right) = - 2 x + 14$
for critical point(s)
$f ' \left(x\right) = 0$
$- 2 x + 14 = 0$
$x = 7$ for the only critical point

We can either use our knowledge of quadratics or take the second derivative ($f ' ' \left(x\right) = - 2 \rightarrow$ slope is always decreasing) to observe that this point is a maximum.

Therefore $f \left(x\right)$ has a maximum when $x = 7$
$f \left(7\right) = - \left(49\right) + 98 - 48 = 1$

So the range of $f \left(x\right)$ is $\left[- \infty , + 1\right]$