Expert: Sherman D. - 4/23/2004

Question
Well, I have lots of questions...here it goes!

1. GIven y = -2x^2 - 6x + 7 How do I find the maximum or minimum value of y and then determine the value of x at which the max or min occurs?

2. How do I use the graph of y = x^3 - 2x^2 - 5x +6  to factor the polynomial without the "y"? And then solve the inequality of it >0?

3. How do I graph y = 2x-1/x-3?  And then, how do I find the vertical and horizontal asymptotes?

Please show all your work...Thanks.


Answer
1.)
y = -2x^2 - 6x + 7
x = -b/2a
x = -(-6)/(2*-2)
x = 6/-4
x = -3/2

y = -2(-3/2)^2 - 6(-3/2) + 7
y = -2(9/4) + (18/2) + 7
y = (-18/4) + 9 + 7
y = (-9/2) + 16
y = (-9/2) + (32/2)
y = (-9 + 32)/2
y = (23/2)

Vertex is at ((-3/2),(23/2))
Since the highest power is negative, this would be a maximum.

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2.)
If you go to www.calculator.com, click on Graphing, and type in x^3 - 2x^2 - 5x + 6, you will get x = -2, 1, and 3, so the factoring would be

(x + 2)(x - 1)(x - 3)

3.)
For a graph, same above, and type in (2x - 1)/(x - 3)

vertical asymptotes are the easiest to find.

Take the doniminator, and make it equal to zero, and solve, and that is the vertical asymptote, so

Vertical Asymptotes is x = 3

To find the horzizontal assymptotes, just switch the x's with y's, and solve for y, and you get

y = 2x - 1 / x - 3
x = 2y - 1 / y - 3
xy - 3x = 2y - x
xy - 2y = x + 3x
y(x - 2) = 4x
y = (4x)/(x - 2)

Horizontal Asymptote is y = 2