Units 1 Questions thread

[superflya]

yes u differentiate and let it equal zero. to find ur y coordinate sub the x value/s back into ur original equation. to test for the nature of the point use the box thingy, dont think u wouldve learnt second derivatives.

[cltf]

yes u differentiate and let it equal zero. to find ur y coordinate sub the x value/s back into ur original equation. to test for the nature of the point use the box thingy, dont think u wouldve learnt second derivatives.

haven't learnt it yet.

so let ? then.. but how would you find the y coordinate? 

[davidle_10]

So say you have the equation y=ax^3+bx^2+cx+d (cubic)
Firstly you have to differentiate so: dy/dx  =3ax^2  +  2bx + c  (differentiation of polynomials)
Then when the derivative=0 (stationary point), solve for x
After you have found the x value(s) you can sub them back into the original equation,
y=ax^3+bx^2+cx+d, to find the corresponding y coordinate.

[cltf]

so
or does dy/dx have an actual rule?

[davidle_10]

dy/dx (called the derivative or gradient function) is essentially the gradient of an equation at any given point. There are rules in determining the derivative of a function. For a polynomial, x^p, the derivative is px^p-1. In the case of y=ax^3+bx^2+cx+d,    dy/dx= 3ax^(3-2)+2bx^(2-1)+cx^(1-1)
                                                  = 3ax^2+2bx+c
But unless your class hasn't started doing differentiation yet, I would just use the calculator to find the point. If you are using the TI-nspire then you will need to press menu,trace,graph trace. Then just move the cursor to the location of the maximum and minimum turning points and the coordinates will show up.

[cltf]

dy/dx (called the derivative or gradient function) is essentially the gradient of an equation at any given point. There are rules in determining the derivative of a function. For a polynomial, x^p, the derivative is px^p-1. In the case of y=ax^3+bx^2+cx+d,    dy/dx= 3ax^(3-2)+2bx^(2-1)+cx^(1-1)
                                                  = 3ax^2+2bx+c
But unless your class hasn't started doing differentiation yet, I would just use the calculator to find the point. If you are using the TI-nspire then you will need to press menu,trace,graph trace. Then just move the cursor to the location of the maximum and minimum turning points and the coordinates will show up.

if reference to the bottom part, the problem im facing is that the points don't go to 2dp. or at not actual enough. so what do i do?

[the.watchman]

Erm ... guys, this is MM unit 1
And anyway cltf, none of the (CGS) 1/2 classes have done differentiation yet, dont worry

For this, you would need to use a calculator for finding local max and min

[the.watchman]

Sorry for the double post, but you could use the fmin() and fmax() functions on your Ti-Nspire, if you restrict the domain

[the.watchman]

Check it in the catalog, but it's fmin(FUNC., VARIABLE)
So for example:

gives

[the.watchman]

That's because the min of is 0, when x = -2 or 2, check the graph

Oh, and yes that's the symbol, it means "given that"

[cltf]

does any one have MME u1&2 semester one - calculator trial exams?

[cltf]

if a turning point touches the x axis is that considered an intercept? if so why?

[luken93]

yes, discriminant = 0

edit: anything that touches/cut the x-axis is a root.

In other words,
Two solutions/intercepts (Cutting the line)
One Solution/Intercept (Touching the line, aka tangent to the line)
No Solutions/Intercepts (Does not touch the line)

[TrueTears]

if a turning point touches the x axis is that considered an intercept? if so why?

it's a matter of definition which is not too important in mathematics, just define what u need but anyways for the sake for completeness, a quote from wikipedia:

"a root (or a zero) of a real-, complex- or generally vector-valued function ƒ is a member x of the domain of ƒ such that ƒ(x) vanishes at x, an alternative name for the root in this context is the x-intercept."

thus if a tp touches the x axis it is an intercept.

[luken93]

okay, heres  one, it was on our exam today, and i'm not sure if i got it, so if anyone wants to have a go...

A circle has its centre at (5 , -1) and a radius of 5
a) Show that ( 8 , 3 ) is a point on the circle

         y= mx is tangent to the circle
b) Find the value(s) of m which makes the line y = mx tangent to the circle

i) Find the value(s) of m which crosses the circle twice

ii) Find the value(s) of m which does not touch the circle at all