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

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Answer 1 · 2015-04-08T21:44:31

Answer: $=> y <= 1$

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

So, letting $y= f(x)$

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

$=> x^2 - 4x + 3 +y = 0$

For real $x$ , $b^2 - 4ac >=0$ For the quadratic $ax^2 + bx + c = 0$

Applying this to our equation,

$=> (-4)^2 - 4(1)(3 + y) >= 0$

$=> 16 - 12 - 4y>= 0$

$=> 4 >= 4y$

$=> y <= 1$

In other terms $(-oo, 1]$

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