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]#