# Math Help - CURVE SKETCHING question

1. ## CURVE SKETCHING question

for the function f(x) = (x-4)(x+1)(x-1)

the domain and range is it x cannot be 4, -1, or 1??

what is "real zeros"??

and how do u figure the y-intercept...do u just plug 0 into all the X's?

what is the end behaviour?

how do we figure out the symmetry

and the number of turning points..is it 2?

and how do u figure out the approximate coordinates of any local maximum points or local minimum points

THANKS SO MUCH!!

and if u don't mind..can u show me a very brief sketch if its possible?

2. Originally Posted by calchelp
for the function f(x) = (x-4)(x+1)(x-1)

the domain and range is it x cannot be 4, -1, or 1??
The domain is all of the real numbers $\mathbb{R}=(-\infty,\infty)$ as it
is defined everywhere.

The range is also all of the reals since this is a cubic with real coeficients
every equation of the form $f(x)=r$ for any $r \in \mathbb{R}$ has at least one real root.

what is "real zeros"??
a "real zero", is a real number $r$ such that $f(r)=0$

and how do u figure the y-intercept...do u just plug 0 into all the X's?
yes

what is the end behaviour?
This is how the function behaves as $x \to \infty$and $x \to -\infty$. This function is a cubic so as $x \to \infty,\ f(x) \to \infty$ and as $x \to -\infty,\ f(x) \to -\infty$

how do we figure out the symmetry
Don't know what you mean here.

and the number of turning points..is it 2?
There must be at least one turning point between each pair of adjacent roots
and there can be no more than $n-1$ turning points for a polynomial function of degree $n$, so yes there are 2 turning points.

and how do u figure out the approximate coordinates of any local maximum points or local minimum points
Depends on what you mean by approximate. I would start with them at the mid-points between consecutive roots.

RonL

3. ## Follow Up Questions On Curve Sketching!!!!!

i still dont understand the real zeros and end behaviour thing, could you show me how its done

and i meant to ask if thers any symmetry, and how do you know?

also, how do u find the local max and min for this graph

4. Originally Posted by calchelp
i still dont understand the real zeros and end behaviour thing, could you show me how its done
The zeros of a function are the solutions of f(x)=0. In this case
you have a cubic in factorised form, f(x)=(x-4)(x+1)(x-1), so the
roots are 4, -1, and 1, as these values each make one of the factors
zero.

The end behaviour is derived by looking for what happens when |x| becomes
large. Here we have a cubic which is of the form:

f(x) = x^3 + some quadratic

Now when |x| becomes large |x^3| dominates and we can ignore the
quadratic and so for large |x|:

f(x) ~ x^3

which determines the end behaviour.

and i meant to ask if thers any symmetry, and how do you know?
There is a point about which a cubic has a symmetry, but finding it
is too fiddly for me to do here. Essentialy your cubic is a translated
copy of the cubic y=x(x+a)(x-a) for some a, which has 180 degree
rotational symmetry about (0,0), so our cubic has such a symmetry
about the image of (0,0) under the same translation.

also, how do u find the local max and min for this graph
Have you done any calculus? Then the local max and min are solutions
of f'(x)=0.

RonL

5. wow thanks for all that help . much needed