Range of Rational Functions...3

[nycmathguy]

For f(x) = (2x^2 - 3x + 4)/(x^2 + 4x - 5), find the range algebraically. Also, graph the function and find the range by looking at the graph of f.

 

[MathLover1]

f(x) = (2x^2 - 3x + 4)/(x^2 + 4x - 5)

Range of :
f(x) = (2x^2 - 3x + 4)/(x^2 + 4x - 5)
the set of values of the dependent variable for which a function is defined

Rewrite as
y= (2x^2 - 3x + 4)/(x^2 + 4x - 5)
y(x^2 + 4x - 5)= 2x^2 - 3x + 4
yx^2 + 4xy - 5y= 2x^2 - 3x + 4
yx^2 + 4xy - 5y-2x^2 +3x - 4=0
(yx^2 -2x^2)+ (4xy +3x)- (5y + 4)=0
(y -2)x^2+ (4y +3)x- (5y + 4)=0

the range is the set of y for which the discriminant is greater or equal to zero
b^2-4ac>=0
(4y +3)^2-4(y -2)(- (5y + 4))>=0
(4y +3)^2-4(y -2)(- (5y + 4))>=0
16y^2 + 24y + 9-(-20 y^2 + 24 y + 32)>=0
16y^2 + 24y + 9+20 y^2 - 24y - 32>=0
36y^2 - 23>=0
y^2 >= 23/36
y >= sqrt(23/36)
y >= sqrt(23)/6

range:
y >= sqrt(23)/6 or y <= -sqrt(23)/6

 

[nycmathguy]

f(x) = (2x^2 - 3x + 4)/(x^2 + 4x - 5)

View attachment 492


Range of :
f(x) = (2x^2 - 3x + 4)/(x^2 + 4x - 5)
the set of values of the dependent variable for which a function is defined

Rewrite as
y= (2x^2 - 3x + 4)/(x^2 + 4x - 5)
y(x^2 + 4x - 5)= 2x^2 - 3x + 4
yx^2 + 4xy - 5y= 2x^2 - 3x + 4
yx^2 + 4xy - 5y-2x^2 +3x - 4=0
(yx^2 -2x^2)+ (4xy +3x)- (5y + 4)=0
(y -2)x^2+ (4y +3)x- (5y + 4)=0

the range is the set of y for which the discriminant is greater or equal to zero
b^2-4ac>=0
(4y +3)^2-4(y -2)(- (5y + 4))>=0
(4y +3)^2-4(y -2)(- (5y + 4))>=0
16y^2 + 24y + 9-(-20 y^2 + 24 y + 32)>=0
16y^2 + 24y + 9+20 y^2 - 24y - 32>=0
36y^2 - 23>=0
y^2 >= 23/36
y >= sqrt(23/36)
y >= sqrt(23)/6

range:
y >= sqrt(23)/6 or y <= -sqrt(23)/6

Sorry but still unclear.

 

[MathLover1]

to find the range you need to rewrite f(x) = (2x^2 - 3x + 4)/(x^2 + 4x - 5) as quadratic function ax^2+bx+c=0

in your case we got (y -2)x^2+ (4y +3)x- (5y + 4)=0 => a=(y -2), b=(4y +3), and c=- (5y + 4)

then use discriminant to find value(s) for what y>=0

 

[nycmathguy]

to find the range you need to rewrite f(x) = (2x^2 - 3x + 4)/(x^2 + 4x - 5) as quadratic function ax^2+bx+c=0

in your case we got (y -2)x^2+ (4y +3)x- (5y + 4)=0 => a=(y -2), b=(4y +3), and c=- (5y + 4)

then use discriminant to find value(s) for what y>=0

This requires much practice. I will post many more FIND THE RANGE of functions when time allows.