SOLUTION: 1)Give the equation of the horizontal and the vertical asymptote of
R(x)=x^3-1/x^3+1
2)Given the quadratic function F(x)=-x^2-6x-3, is the vertex a MAXIMUM or a MINIMUM for t
Algebra.Com
Question 154431: 1)Give the equation of the horizontal and the vertical asymptote of
R(x)=x^3-1/x^3+1
2)Given the quadratic function F(x)=-x^2-6x-3, is the vertex a MAXIMUM or a MINIMUM for the function and what is the value of that MAXIMUM or the MINIMUM?(note the value of the max or min is the value of the function,not the coordinate of the vertex)
3) Given the function F(x)=x-1/x+1 Find the domain and range of F in interval notation.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1)Give the equation of the horizontal and the vertical asymptote of
R(x)=x^3-1/x^3+1 = [(x-1)(x^2+x+1)] / [(x+1)(x^2+x+1)]
horizontal asymptote: y = x^3/x^3 = 1
vertical asymptote: x = -1
---------------------------------
2)Given the quadratic function F(x)=-x^2-6x-3, is the vertex a MAXIMUM or a MINIMUM for the function and what is the value of that MAXIMUM or the MINIMUM?(note the value of the max or min is the value of the function,not the coordinate of the vertex)
---------
Since the coefficient of x^2 is negative the parabola will open downward
so the vertex will be a maximum.
The max occurs when x = -b/2a = --6/(2*-1) = 6/-2 = -3
f(-3) = -(-3)^2 - 6(-3) -3 = -9+18-3 = 6
Vertex: (-3,6)
-----------------------------------
3) Given the function F(x)=x-1/x+1 Find the domain and range of F in interval notation.
Domain: All Real Numbers except x = -1 because the vertical asymptote is x=-1.
Range: All Real Numbers except y = 1 because the horizontal asymptote is y=1.
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Cheers,
Stan H.
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