Mathematics Q?

Question

1. The parabola y=-1/4(x-r)(x-s) intersects the axes at three pts: (-k,0) and (3k,0) on x-axis, and (0,3k) on y-axis. The vertex
of this parabola is the point V. Determine the value of k and the coordinates of V.

2. Prove that (secx)^6 = (tanx)^6 + 3(tanx secx)^2 + 1.

Answer 1

1) The parabola y = -1/4(x - r)(x - s) intersects the axes at three pts: (-k,0) and (3k,0) on x-axis, and (0,3k) on y-axis.
The vertex of this parabola is the point V. Determine the value of k and the coordinates of V.

The x-intercepts occur at

x = r, s
So
r = -k
s = 3k

Plugging into the equation we have:

y = -1/4(x - r)(x - s)
y = -1/4(x + k)(x - 3k)
y = -1/4(x² - 2kx - 3k²) = -x²/4 + kx/2 + 3k²/4

Plug in x = 0 to find the y-intercept.

y = 3k²/4 = 3k
k/4 = 1
k = 4

x-intercepts are:
r = -k = -4
s = 3k = 12

y-intercept is:
3k = 12

The equation of the parabola is:

y = -1/4(x² - 2kx - 3k²)
y = -1/4(x² - 2*4x - 3*4²)
-4y = x² - 8x - 48
-4y + 48 = x² - 8x
-4y + 48 + 16 = x² - 8x + 16
-4y + 64 = (x - 4)²
-4(y - 16) = (x - 4)²

The vertex is (4, 16).
____________

2. Prove that (secx)^6 = (tanx)^6 + 3(tanx secx)^2 + 1.

Remember the identity:
sec²x = tan²x + 1

(sec x)^6 = (sec²x)³ = [(tan²x) + 1]³

= (tan x)^6 + 3(tan x)^4 + 3tan²x + 1

= (tan x)^6 + 3tan²x(tan²x + 1) + 1

= (tan x)^6 + 3tan²x(sec²x) + 1

= (tan x)^6 + 3(tanx secx)² + 1

Answer 2

y = (1/4)(x + k)(x - 3k)
4y = x^2 - 2kx - 3k^2
12k = - 3k^2
k^2 + 4k = 0
k(k + 4) = 0
k = 0 is trivial, so
k = - 4

y = (1/4)(x^2 + 8x) - 12
y = (1/4)(x^2 + 8x + 16) - 4 - 12
y = (1/4)(x + 4)^2 - 16
V = (- 4, - 16)

edit: Northstar is correct. I missed the - sign.

(secx)^6 =? (tanx)^6 + 3(tanxsecx)^2 + 1
(tan^2x + 1)^3 =? (tanx)^6 + 3(tanxsecx)^2 + 1
tan^6x + 3tan^4x + 3tan^2x + 1 =? (tanx)^6 + 3(tanxsecx)^2 + 1
tan^6x + 3tan^2x(sec^2x - 1) + 3tan^2x + 1 =? (tanx)^6 + 3(tanxsecx)^2 + 1
tan^6x + 3tan^2xsec^2x - 3tan^2x + 3tan^2x + 1 =? (tanx)^6 + 3(tanxsecx)^2 + 1
(tanx)^6 + 3(tanxsecx)^2 + 1 =
(tanx)^6 + 3(tanxsecx)^2 + 1

Answer 3

80 pence cases a million/4 skill 80 cases a million then divided through 4. it extremely is 20. so as it really is what he spends. He has left what he had minus what he spends so 80 - 20 = 60 If 24 out of sixty 4 do no longer placed on glasses so sixty 4 -24 = 40 do placed on glasses. 40 divided through sixty 4 is the fraction =40/sixty 4 yet 8 is an elementary aspect of sixty 4 and 40 so dividing the proper and bottom through 8 supplies = 5/8