To find the stationary point for a parabola, use the fact that this occurs at the vertex, which is on the axis of symmetry, or x = -b/(2a).
This is in the subject content of an exam I plan to sit but I don't have my text book yet so I am working from free resources online, but I have never seen this before and I can't find anything like it by searching google, hopefully someone here can tell me where to find more info on this.
"Using quadratic theory (not calculus) to find the possible values of the function and the coordinates of the maximum or minimum points on the graph.
Eg for y = x^{2}+2/x^{2}-4x, if y = k then x^{2}+2 =kx^{2} - 4kx,
Which has real roots if 16k^{2} + 8k - 8 >/= 0, ie if k </= -1 or k >/= 1/2; (Greater than or equals: >/=)
Stationary points are (1, -1) and (-2, 1/2)."
Thanks,
-Jake
you can't really tackle "general rational functions" with just "quadratic theory".
however, if f(x) = P(x)/Q(x), where P(x) and Q(x) are polynomials of degree < 3, then one can use what one knows about "quadratic polynomials" (parabolas).
when we set y = k, what we are doing is trying to find (local) maximum and minimum values for k (which is y, the value of f(x)).
using the quadratic theory, we then look for the x's that give those values for y:
y = -1 --> x^{2} + 2 = 4x - x^{2} --> 2x^{2} - 4x + 2 = 0 --> x^{2} - 2x + 1 = 0 --> x = 1.
so (1,-1) is a stationary point.
y = 1/2 --> 2x^{2} + 4 = x^{2} - 4x --> x^{2} + 4x + 4 = 0 --> x = -2.
so (-2,1/2) is another stationary point.
**********
so far, this only tells us f has a "peak" or a "valley" at those two points (or possibly, an inflection point). since the denominator because undefined at x = 0, and x = 4, we want to look at how f behaves on the following regions:
-∞ < x < -2
-2 < x < 0
0 < x < 1
1 < x < 4
4 < x < ∞
"near" -∞, (very large negative x) f(x) is near 1. f(-2) = 1/2, indicating f is falling on this interval.
on the left side of 0, f is positive (look at the denominator), and on the right side, f is negative. this tells us f has a "valley" at x = -2, and a "peak" at x = 1.
for 1 < x < 4, f remains negative, so falls sharply near x = 4.
for x > 4, f is positive again, so starts out "high", then gradually settles to near 1, as x becomes large.
That all make sense but how do you decide on values for k in the first place?
It says that x^{2}+2 =kx^{2} - 4kx has real roots if 16k^{2} + 8k - 8 > 0 or = 0
But I don't understand how they get from x^{2}+2 =kx^{2} - 4kx to a quadratic for k.