Please help! Mental block!
Solve the following inequation, expressing your answer in set notation:
x^2 - x - 5 < 0
and...
What is the equation of the porabola that goes through the points:
(0,0)
(1,-2)
(4,4)
please help~ i'm desperate!
Hello, georgiaaa!
Think of it as: .y .= .x² - x - 5Solve the following inequation, expressing your answer in set notation:
. . x² - x - 5 .< .0
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
The general form of a parabola is: .y .= .ax² + bx + cWhat is the equation of the parabola that goes through the points:
. . (0,0), (1,-2), (4,4)
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