Thread: Sketching graphs

Can anyone help me in sketching a graph of N against t from this?

t² +


Maybe then i can figure out how to then sketch a graph of this:

t

If anyone can help i'd be really grateful, thankyou

Hey bensho92.

Try evaluating (i.e. calculating) various points for t and then fit a line through the points in a smooth way.

Yeah, that solution's a good one - it works for any function.

For your first N(t), most students learn that the graphs of equations of the form are parabolas - opening in the up direction if a is positive and the down direction if a is negative.

You can also easily find two points on the x-axis.

- Hollywood

Originally Posted by chiro

Try evaluating (i.e. calculating) various points for t and then fit a line through the points in a smooth way.

How do i do this? Thanks

Basically pick a value of t (like say t = 1) and plug it into the formula to get the y co-ordinate.

Do this for enough values of t and you will get a lot of dots which you can connect up and draw a smooth curve through if you have enough dots on your graph paper.

Does it look something like this?

Originally Posted by bensho92

Does it look something like this?

no. as hollywood stated, the graph is a parabola that opens downward.

start by locating the zeros





,

vertex (maximum in this case) of the parabola is at

the first thing to do is sketch some "easy points" like (0,N(0)), (1,N(1)) and (-1,N(-1)).

this gives us the points (-1,-36/25), (0,0) and (1,24/25) without much effort.

the second thing to do is imagine what happens if |t| gets very big. in this case, the t² will eventually "dominate", and since t² is always positive and we are multiplying it by -6/25, at the "far ends" of the real line, N(t) << 0 (very negative).

since our 3 points we found go "up and up" and N(t) is very negative, for large t > 0, somewhere N must "turn around" and go back down.

for polynomial functions there is an easy rule:

if p(t) = a₀ + a₁t +...+ a_ntⁿ

there is another polynomial (called the derivative of p):

p'(t) = a₁ + 2a₂t +....+ (n-1)a_nt^n-1

that keeps track of "which way p is headed".

for our original N(t), this is:

N'(t) = 6/25 + (-12/25)t

if N'(t) > 0, N is "going up"

if N'(t) = 0, N is "leveling out"

if N'(t) < 0, N is "going down"

this tells us (for your first N):

if (12/25)t < 6/5, N is going up

if (12/25)t = 6/5, N is leveling out

if (12/25)t > 6/5, N is going down

the "critical point" for this particular N, is t = (6/5)(25/12) = 5/2 <---this is the "top of the mountain".

analyzing where N(t) = 0 CAN be useful, but not all polynomials factor very easily.

in this case we see that N(0) = 0, and N(5) = 0, which tells you somewhere between t = 1 and t = 5, N must have been doin' some downward travel.

*********

let's see what we get with N(t) = (-1/25)(t⁵)(t+4).

easy points to figure out:

N(0) = 0
N(-1) = (-1/25)(-1)(3) = 3/25
N(1) = (-1/25)(1)(5) = -1/5

since this is already factored, we can find the places N crosses the t-axis easily:

at t = 0, and t = -4 (looks like we have at least one "hump" between t = -4, and t = 0, since N(-1) > 0).

as |t| gets big, N(t) will be large and negative for negative t's, and large and negative for positive t's (starts low on the left coming up....eventually goes further down on the far right).

to find N'(t), we have to expand N(t):

N(t) = (-1/25)t⁶ - (4/25)t⁵

N'(t) = (-6/25)t⁵ - (4/5)t⁴.

we can factor this as:

N'(t) = (-2/5)(t⁴)((3/5)t + 2)

we now have 2 places where N "levels out": at t = 0, and t = -10/3 <---this must be the "top of the hump" we suspected earlier. t = -10/3 would be a good point to plot.

since N(0) = 0, but N never crosses the t-axis again, it must just level flat for a moment, then turn back down again (like the middle of the letter S).

so N(t) looks like an upside-down U, with a little bump on it around t = -10/3.

for your plotting convenience:

N(-10/3) = (-1/25)(-1000/27)(-2/3) = 8000/729 ~ 10.97

Originally Posted by bensho92

Can anyone help me in sketching a graph of N against t from this?

t² +

That's a parabola, opening downward because of the "-" on the coefficient of . And it is easy to see that if t= 0, N= 0. I would also "complete the square". Write this as
Now, you know that . The first two terms of that will be the same as if which means that a= 1/2 and . Adding and subtracting 1/4, .

The point of doing all that we can see that when t= 1/2, t- 1/2= 0 so N(1/2)= 3/10 and if t is any other number so N(t) is less than 3/10. That is, (1/2, 3/10) is the vertex of the parabola.

Maybe then i can figure out how to then sketch a graph of this:

t

If anyone can help i'd be really grateful, thankyou

This doesn't really have anything to do with the first one. Note that N(0)= N(5)= N(-4)= 0. If t> 5, all three factors are positive but the leading coefficient is negative so N(t) is negative and the graph goes down as t goes to infinity. On the other hand, if x< -4, all three factors are negative so, multiplying by -1/25, N(t) is positive and the graph goes up as t goes to -infinity. Putting all of that together, the graph come down from the left, passes through (-4, 0), becoming negative, but then turns back up, passing through (0, 0), becomes positive again, but then turns back down, passing through (5, 0) and then down to the right.

After you take Calculus, you will be able to determine "local max and min" for the function and so decide exactly where that turning occurs.

the reason i introduced the derivative in post #8, is that, for polynomials, you don't HAVE to use calculus. it's just another polynomial you compute using a certain rule (where this rule COMES from is another story).

as far as my assertion that N'(t) keeps track of where N(t) is headed...well, he'll just have to take that on faith, for now. this is not that uncommon, we don't teach school-children that the integers are a unique factorization domain before telling them about prime numbers.