How to Construct a Polynomial Function Given Its Graph



Have you ever wondered how to construct
your own polynomial equations to satisfy certain conditions? Well, I used to wonder [about] that
when I was in junior to middle high school. And I'­ve since discovered
a couple of ways of doing this. The one I'­m going to show you in this video
is how to use the roots and the y-intercept. In a subsequent video I'­m going to show you
how to use a table of values to generate a polynomial equation —
a rather more sophisticated deal. But, for the moment, we'­re going to look at constructing
polynomial equations using roots and y-intercepts. I have behind me the graph
of a cubic equation that I'­ve just drawn. I actually like hand drawing equations —
one gets a better feel for things. You'­ll notice that it has a root at +3,
a root and -1 and a root at -5. And let'­s say the intercept is at -4. I'­ve just made these numbers up
"on the spot," and the question is,
"How do I construct an equation, a polynomial equation,
that will produce this graph?" Well, we work in reverse
[… along the three …] following the three procedures
that we used in the previous videos. That is, we look at the roots,
we look at the leading coefficient — and we look to see whether
the leading coefficient is positive or negative
and we look at its size. Now, to generate the roots is quite easy. We simply need factors that will give us
roots in these positions. You might remember that, to find the roots,
we have to set the y value equal to zero. And that means that any of these factors
[could be] zero. So, to have a zero when x is 3,
we need a factor of (x-3) because 3 take away 3 is zero. To have a root at -1
we need a factor of (x+1); and to have a root at -5,
we need a factor of (x+5) because -5+5 is zero. That — that structure —
takes care of all the roots. It is as simple as that! Now, the leading coefficient. You might remember that any polynomial
that goes up to the right has a positive coefficient. That means we expect to have
a positive number there. Now, of course, we'­re not going
to write the plus sign in. That'­s rather redundant. The final question is, "[What number belongs] …
What size is our leading coefficient?" And that'­s when we need to use
information about the y-intercept. Remember, we already have these in place —
and the fact that we have [… three terms …] three factors multiplied together
gives us this cubic shape, and the positive coefficient
gives us the positive gradient as x increases, so this is our last bit of the puzzle. Now, this happens on the y-axis
(which is where x is zero). Remember that it'­s the y-value
that keeps changing; it'­s the x-value that remains at zero because you haven'­t gone to the right
or the left along the x-axis; you'­ve remained at the zero position. So, what we do is, we substitute
zero [for] x in here and we substitute -4 for the y-value
(because it lies on our graph) in here and we'­ll perform a little bit of algebra
to find the missing coefficient. So, -4 = a(0+5)(0+1)(0-3). Let'­s solve this quickly —
"a" times 5 times 1 times -3 which is "a" times … [well, actually,
let'­s put the coefficient in front of it] … 5 times 1 times -3 is -15.
And, if -4 equals that, then "a" is going to be -4 on -15
which is 4/15. And this means that our final cubic equation
is y = 4/15 … it'­s positive, as we expected! …
times (x+5)(x+1)(x-3). There it is, without a great deal of polish —
but that'­s the process. I'­m going to do it twice more
in this video. I'­m going to speed the process up
because you can stop it at any stage and review sections of the video
if you wish. We'­ll study a quadratic next,
a very simple one. Just choose whichever roots.
Let'­s have one at -2 and one at +4. And, let'­s be a bit adventurous
and make it an upside-down parabola, and give it a y-intercept of +5. I chose that because it'­s odd
and the roots are even — we might find something interesting. Let'­s go through the same procedure. To have the roots in these two places
we need to have [… one root or …] one factor of (x-4)
to have a root at x [equals] +4. And another factor [of] (x+2)
so that x=-2 will make that zero. And that takes care of our two roots. We let the coefficient be "a," and we now substitute the coordinates
for this point … which happens to be (0,5) because x is worth zero (remember?)
on the y-axis and the y-value is worth 5. So, y is worth 5, x is worth zero …
I'­ll put in all the steps … 0-4 is -4, 0+2 is +2 … that'­s -8.
And now we have -8a = 5 and "a" must be -5/8. Notice, we have a negative coefficient —
I'­ll put that back in place of the "a" up here. Notice that we basically had the equation here and all of this work
is just to find that coefficient, to get it [the graph]
to go through the y-intercept. It'­s a negative coefficient which means
the graph is going downwards to the right, as we expected. And the 5/8 is just the number we need to use
to adjust it so it goes through that point. And there are our two roots from the two factors.
A fairly simple procedure. So, let'­s do one last example
which is a little bit more complicated. We'­re going to draw
a rather more complicated polynomial. Let'­s have a double root, a triple —
how'­s that? We'­ll give them some numbers —
let'­s say that'­s at 2, 4, let'­s say that'­s at 6, and that'­s at 8
— we'­re going to get some large numbers here — I just hope my mind is up to it —
so there we go, and we'­ll choose a y-value,
a y-intercept of -60. This is being a bit adventurous,
but let'­s see what happens. Let'­s organize our roots first. We have four roots;
two of them are normal roots, but we have a triple root here
and a double root here. You might remember from a previous video
that triple roots have the curve pass through them
in a kind of "S" shape, and double roots have the curve
"bounce" off the axis, where the root is. So, we can write our root at -4,
or our factor would be (x+4), and we'­re going to have that one twice. Our factor at x=2 is going to be (x-2). We'­re going to have that three times —
it'­s a triple root. Our factor here is going to be (x-6),
and our factor here is going to be (x-8). That has now got our polynomial
going through all these points. I'­m expecting a positive coefficient.
I hope you are, too, because I notice the graph
is going up to the right. But, we'­re going to use
our general technique to find the coefficient
by substituting our y-intercept. So, when x is worth zero
the y-value is going to be -60. So, let'­s check this out.
-60 is "a" lots of — and, because I don'­t have much room,
I'­m going to substitute zero immediately — 0+4 is 4, 0-2 is -2, -6, -8. So, we'­re going to have "a" times 16 times -8
(and I'­ll multiply these), six eights are forty-eight,
and the two minuses make a plus. Now, this has the potential
to hurt one'­s head a little bit but we have two minus signs,
so I can remove those on the next line. And I want a number that divides into 60
that also divides into these numbers. So, let'­s start by dividing by 4. Fours into sixty is fifteen …
and fours into, let'­s say, 48 is 12. Now 3 will divide out, [… sorry, that'­s a "+" now,
'cause we got rid of our two minus signs …] and 3 into fifteen is 5,
3 into twelve is 4. There'­s nothing else [that] divides out. So, "a" is going to be worth
5 over 16 by 8 by 4. Now, I didn'­t plan this,
but these are all powers of 2. This is 2 to the fourth power,
2 cubed and 2 squared … so, four and three and two is nine
and 2 to the ninth power is 512 … and that'­s as simple as we can get it. So, our final polynomial will be
5/512(x+4)^2(x-2)^3(x-6)(x-8) And that is a rather complex polynomial
that we'­ve generated just by knowing its four roots
and its y-intercept. I'­m going to create a worksheet for you
to practise this skill (it'­ll have a few more instructions on it),
but I hope this is helpful to you. And in the next video
[it was going to be this one] … but in the next video I'­m going to explain
how you can use this principle of roots or zeros
to generate graphs that have bits in different places
on the graph paper and even generate the words "I LOVE YOU"
from one equation! So, do look at that video, please. Thank you.

24 thoughts on “How to Construct a Polynomial Function Given Its Graph

  1. On your third example, where do you get y equal to -60? Is the test point not (0, -6)? That would make y equal to -6?… yes? no? Makes a bit of a difference in the answer. By my calculations, a = 1/1024, not your answer of 5/512. Please explain.

  2. I tried creating equations for multiple graphs that i have randomly drawn by hand. And tried to recreate those graphs in excel using the equations. Wow…It turned out right and it gave me a lot of confidence. You sir have made maths interesting for me.

  3. Thank goodness for this video, it seriously helped me to understand how to get polynomial functions from a graph! 🙂

  4. Thanks, it was helpful. i would recommend useing different colored pencils though, it makes it easier to differentiate between the various components.

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