# How do you find the range of f(x)=-x^2+4x-3?

Apr 8, 2015

Answer: $\implies y \le 1$

Solution:
The range of this function is the set of the values of $y$ that the $f$ can take,

So, letting $y = f \left(x\right)$

$y = - {x}^{2} + 4 x - 3$

$\implies {x}^{2} - 4 x + 3 + y = 0$

For real $x$ , ${b}^{2} - 4 a c \ge 0$ For the quadratic $a {x}^{2} + b x + c = 0$

Applying this to our equation,

$\implies {\left(- 4\right)}^{2} - 4 \left(1\right) \left(3 + y\right) \ge 0$

$\implies 16 - 12 - 4 y \ge 0$

$\implies 4 \ge 4 y$

$\implies y \le 1$

In other terms $\left(- \infty , 1\right]$