The first question is quadratic equations and functions, and the second question is parabolas. Please can someone help me, I've been at these two problems for over an hour and I still cannot figure them out. THANK YOU!!!
2007-11-29 04:47:30 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Question 1
(- 4) x (- 4) x (- 4) = 16 x (- 4) = (- 64)
x = (- 4)
Question 2
y = x² - 5x
y = (x² - 5x +25/4) - 25/4
y = (x - 5/2)² - 25/4
Axis of symmetry is x = 5/2
You deserve a rest!
2007-12-02 23:26:43 · answer #1 · answered by Como 7 · 0⤊ 0⤋
x^3=-64
=> x = - 4
x = 5/2 is the axis of symmetry for the parabola as worked below.
y=x^2-5x
=> y = (x - 5/2)^2 - 25/4
=> y + 25/4 = (x - 5/2)^2
Shifting origin to (5/2, -25/4), new coordinates (x', y') will be given by
x' = x - 5/2 and y' = y + 25/4
=> y' = x'^2
This is a parabola for which y'-axis is the axis of symmetry whose equation is x' = 0.
With respect to old coordinate system it is x - 5/2 = 0 or x = 5/2
2007-11-29 12:55:53 · answer #2 · answered by Madhukar 7 · 1⤊ 0⤋
x^3=-64 => x=-4.
The axis of symmetry of y=x^2-5x is x=5/2.
2007-11-29 12:52:57 · answer #3 · answered by JP 3 · 0⤊ 0⤋
hey! i am more than sure that your first question's (x^3=-64) answer would be -4. (-4) * (-4) = +16. so, 16 * (-4) = -64.
or -4 * -4* -4 = -64. i hope that helps you. as far your second question i cant exactly remember how to do it without a math book. :)
2007-11-29 12:57:01 · answer #4 · answered by Anonymous · 0⤊ 0⤋
x^3=-64
x=-4.
2.y=x^2-5x
y=x^2-2*x*5/2+(5/2)^2-25/4
=(x-5/2)^2-25/4.
the axis of symmetry of the parabola is:x=5/2.ANS.
2007-11-29 13:02:20 · answer #5 · answered by Anonymous · 0⤊ 0⤋
First one is x = -4. Do you need to know how to do it, or just the answer?
2007-11-29 13:06:44 · answer #6 · answered by lpf 2 · 0⤊ 0⤋