H2 Math Qn

Okazaki SG

FireIce SG

ur qn not complete?

wee_ws SG

Hi,

The graph of y = f(x) is symmetrical about the x-axis means that the points (x, f(x)) and (x, -f(x)) lie on the graph. Thanks.

Cheers,
Wen Shih

Okazaki SG

Hi thanks.what about the first part?How do I prove if a function have no stationary point, then the unkown constant in the function would lie in a certain range
Of values?

wee_ws SG

Hi,

Your question is missing some information.

If the graph has no stationary points, we need to consider the equation dy/dx = 0. Typically, we will encounter a quadratic equation and we will then conclude that its discriminant < 0.

Thanks.

Cheers,
Wen Shih

Okazaki SG

Oh ok thanks!I was confuse all the while why discriminant is involved.tyty

frekiwang SG

dy/dx=[(2ax)(x-1)-(ax^2+2)1]/(x-1)^2=(ax^2-2ax+2)/(x-1)^2

When dy/dx=0,

ax^2-2ax+2=0

The curve has no stationary point if this equation has no real roots.

b^2-4ac<0

4a^2-8a<0

4a(a-2)<0

0<a<2

frekiwang SG

At x=a,

a=t^2+3

t=+/- sqrt(a-3)

y = t(x)=+/-sqrt(a-3)a

Which implies for any x=a, y has two possible values symmetrical about x-axis. Therefore the graph is symmetrical about x-axis.

aunty ling SG

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