• Apr 22nd 2007, 05:08 AM
georgiaaa.

x^2 - x - 5 < 0

and...

What is the equation of the porabola that goes through the points:
(0,0)
(1,-2)
(4,4)

• Apr 22nd 2007, 05:56 AM
ThePerfectHacker
Quote:

Originally Posted by georgiaaa.

x^2 - x - 5 < 0

I like to complete the square for tis one.

x^2 -x + 1/4 < 5 + 1/4

(x-1/2)^2 < 21/4

-sqrt(21)/2< (x-1/2) < sqrt(21)/2

(1-sqrt(21))/2 < x < (1+sqrt(21))/2
• Apr 22nd 2007, 06:23 AM
Soroban
Hello, georgiaaa!

Quote:

. . x² - x - 5 .< .0

Think of it as: .y .= .x² - x - 5

We have an up-opening parabola. .When it is negative?
. . When it is below the x-axis.

When does that happen?
. . Let's find the x-intercepts.
. . . . . . . . . . . . . . . . . . . . . . __
. . . . . . . . . . . . . . . . . . 1 ± √21
Quadratic Formula: .x .= . ---------
. . . . . . . . . . . . . . . . . . . . .2

The expression is negative between those two intercepts.
. . . . . . __ . . . . . . . . . . . . . __
. . (1 - √21)/2 .< .x .< .(1 + √21)/2

Quote:

What is the equation of the parabola that goes through the points:
. . (0,0), (1,-2), (4,4)

The general form of a parabola is: .y .= .ax² + bx + c

Plug in the three points:
. . .(0,0): .a·0² + b·0 + c .= .0 . . . c = 0
. . (1,-2): .a·1² + b·1 + 0 .= .-2 . . . .a + .b .= .-2 .[1]
. . .(4,4): .a·4² + b·4 + 0 .= .4 . . . 16a + 4b .= .4 . [2]

Divide [2] by 4: . 4a + b .= .1
. . Subtract [1]: . .a + b .= .-2

. . and we get: .3a = 3 . . a = 1

Substitute into [1]: .1 + b .= .-2 . . b = -3

Therefore, the parabola is: .y .= .x² - 3x