Education Quadratic Graphs

just a quick one that I would really appreciate some help with

Is there any way on a quadratic graph that it can have just one minimum but a value of the power of x is between 1 and 2.

I.e if I have a equation of y = 3 * x^1.7 can that still give me a quadratic graph?

Any quick help is really appreciated

Ocha

 

y = 3 * X^1.7 is a single-valued curve, i.e. every unique point on the x-axis maps to another, unique point on the y-axis. It has no minimum per se, not even a turning point because after you cross into the negative boundary, you end up with complex (aka imaginary) numbers.

So in an answer to your question, no, you can't have a sub-quadratic graph with either a turning point or minima...

 

ok so on that basis the only way a graph could turn and have one minima would be if it had was something along the lines of

y = 3 * x^2

i.e. the 2 being the key?

ocha

 

The only way a y=c*x^p curve will only have 1 absolute minimum is when c>0 and p=2*n with n=1,2,3,4,....

 

Incorrect; imaginary numbers are a result of the power being negative (i.e. y=c*x^p with p<0)

 

No, X^(-2) = 1/X^(2)

 

D'oh my head wasn't on straight

Meant to say complex numbers occur when 0 < |p| < 1 and x < 0

 

To put it simply, the function is discontinuous when x < 0. That means you can't differentiate the point x = 0, and therefore there is no rate of change and no minimum can be mathematically derived (you cannot take the limit of x -> 0-; the rate of change as x approaches zero from the left)

So I guess a proper answer is to say that the lowest perceivable (meaning unproven) point of the function is x = 0, but you cannot mathematically prove it since the x < 0 is undefined. It's a valid function, just no way of proving a minimum using derivatives aside from common sense.

 

only semi-discontinuous for x<0 if the absolute value of the power is between 0 and 1 (results complex numbers) or if the power is 0 (horizontal line)

 

And you're assuming that c > 0. If c=0, it's a straight line along y=0, if c<0, the formula will approach -infinity at both ends and have an absolute maximum.

 

I was trying to put it reasonably simply. You'll end up with a purely imaginary number in this case, but lumping the two together isn't necessarily (imho) a bad thing, but only when you don't need the details.

 

correct as long as the power is p=2*n with n=1,2,3,4,...

other case for an absolute maximum is when c>0 and p=-2*n; absolute minimum also for c<0 and p=-2*n