【科技英语强人团】树的儿子为您解答如下：
a) 分析：要求a,b,c的值,已知有3个条件,我们一一来看：
The graph of f is symmetric with respect to the y-axis.即f（x）的图象关于y轴对称,也就是说f（x）是个偶函数（even function).那么f（-x）=f（x）.将x和-x代入函数表达式f(x)=(ax+b)/(x^2-c),建立等式,化简,得：-a=a,则a=0.于是f（x）=b/(x^2-c) （1）；
lim x-> 2+ f(x)=+00（正无穷）,即当x趋近于2时,函数值趋向于正无穷.由于分子部分是一个定值b,代入x=2不会出现0/0这样的indeterminant form,那么可能使得函数值趋向于正无穷的只有可能是分母为0,即当x=2时,x^2-c=0,解出c=4.于是f（x）=b/(x^2-4）（2）；
f(1)=-2,直接代入（2）解出b=6.
综上,a=0,b=6,c=4,f(x)=6/(x^2-4).
As the graph of f is symmetric with respect to the y-axis,f(x)=f(-x).Plug in x and -x to f(x),we get the equation (ax+b)/(x^2-c)=(-ax+b)/(x^2-c).Simplify it,and we get -a=a.Solve for a:a=0;
Now the numerator part of the function collapes to a single constant b.As lim x-> 2+ f(x) equals positive infinity,the denominator has to be 0 when we plug in x=2.Set up equation,and solve for c--we get c=4;
Finally we plug in 1 for x and the function should be equal to -2.Solve for b and we get b=6.
Thus,a=0,b=6,c=4,and f(x)=6/(x^2-4).
b) Write an equation for each vertical ad each horizontal asymtptote of the graph of f.让你写出f（x）所有水平及垂直渐近线的方程.
分析：用u代替x^2,f（x）就是关于u的反比例函数,此时f(u)=6/(u-4),且u的范围是u>=0.根据函数的平移特性可知,f(u)=6/(u-4)就是f(x)=6/u向右平移4个单位,其垂直渐近线就是u=4,即x^2=4,x=+/-2；由于没有进行垂直平移,f(u)=6/(u-4)的水平渐近线就是f(u)=6/u的水平渐近线,即y=0.
Substitute x^2 with u.Since we know that f(u)=6/(u-4) has vertical asymptote u=4,f(x)=6/(x^2-4) should have two vertical symptotes such that x^2=4--x=+/-2.
Since f(u)=6/(u-4) has identical horizontal asymptote as f(u)=6/u does,f(x) has horizontal asymptote y=0.
Thus, the equations for vertical and horizontal asymptotes of the graph of f are x=-2,x-2 and y=0.
c)注意是偶函数就可以了.自己求一阶导数和二阶导数分别找出monotonity 和 concavity,可以画出图象.