a math qestion!!

Given the curve c: y=(x^2)/(1+x)-4/3,where x not=-1
A)Find the x- and y-intercepts of the curve C.
put x = 0,
y = -4/3
put y = 0,
(x^2)/(1+x) - 4/3 = 0
(x^2)/(1+x) = 4/3
3x^2 = 4(1+x)
3x^2 - 4x - 4 = 0
(3x + 2)(x - 2) = 0
x = -2/3 or x = 2
x-intercepts = -2/3 and 2
y-intercept = -4/3
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B)Find dy/dx and show that d^2y/dx^2 = 2/(1+x)^3
y = (x^2)/(1+x) - 4/3
= x - 1 + 1/(x+1) - 4/3
= x + 1/(x+1) - 7/3
dy/dx
= 1 - 1/(x+1)^2
d^2 y/dx^2
= 2/(x+1)^3
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C)Find the turning point(s) of the curve C
set dy/dx = 0,
1 - 1/(x+1)^2 = 0
(x+1)^2 = 1
x = 0 or x = -2
when x = 0, y = -4/3
when x = -2, y = -16/3
turning points (0, -4/3), (-2, -16/3)
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For each turning point, test whether it is a maximum or a minimum point
d^2 y/dx^2 (at x=0) > 0
d^2 y/dx^2 (at x=-2) < 0
min point = (0, -4/3)
max point = (-2, -16/3)
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In the figure, sketch the curve C for
i)-5<= x <= -1
ii)-1<= x <= 3
呢度就劃唔到個圖比你
大致 x < -1, 向下開口既 U 字, 近 x = -1 時, 負無限大
x > -1, 向上開口既 U 字, cut x-軸 -2/3 and 2, 近 x = -1 時, 正無限大
自己試下劃,
所有資料都幫你搵左

A) For x-intercept, sub y=0
0 = (x^2)/(1+x) - 4/3
4(1+x)=3x^2
3x^2-4x-4=0
(3x+2)(x-2)=0 (by factorization)
--&gt; x-intercepts are x= -2/3 and x=2

For y-intercept, sub x=0
y = 0/(1+0) - 4/3
--&gt; y-intercept is y=-4/3

B) By product rule and chain rule,
dy/dx
= 2x/(1+x) + (-x^2)/(1+x)^2
= [2x(1+x) - x^2] / (1+x)^2
= x(2+x) / (1+x)^2

By Quotient rule,
d^2y/dx^2
= [(2+2x)(1+x)^2 - 2(1+x)(2x+x^2)] / (1+x)^4
= (1+x)[(2+2x)(1+x) - 2(2x+x^2)] / (1+x)^4 &lt;- factorize抽(1+x), then it cancel out with denominator
= [(2+2x)(1+x) - 2(2x+x^2)] / (1+x)^3 之後expand the brackets
= [2+2x+2x+2x^2-4x-2x^2] / (1+x)^3
= 2 / (1+x)^3

C) For turning pts, dy/dx=0.
So put dy/dx = x(2+x) / (1+x)^2 = 0
x(2+x) = 0 &lt;-- since (1+x)^2 is a perfect square =/=0
x=0 , x= -2
When x=0, y=-4/3. When x=-2, y=-16/3
---&gt; Turning points are (0, -4/3) and (-2, -16/3)

Test turning pts using d^2y/dx^2.
When x=0, d^2y/dx^2 = 2/(1+0)^3 = 2 which is &gt; 0
Hence the point (0, -4/3) is a minimum turning point.

When x=-2, d^2y/dx^2 = 2/(1-2)^3 = -2 which is &lt;0
Hence the point (-2, -16/3) is a maximum turning point.

Use the above intercepts and turning points to sketch the curve C.