Question

Write an equation for the 4th degree polynomial graphed below. Use k if your leading coefficient is positive and −k if your leading coefficient is negative. Use y for the dependent variable. You can leave your answer in factored form.

Answer 1

Ho ho ho, lets get this party started
ok so I'm just really excited to use this stuff that I just learned

so

multiplicites
if a root or zero has an even multilicity, the graph bounces on that root
if the root or zero has an odd multiplicty, the graph goes through that root

so
roots are
-1
2
4
multiplicty is how many times it repeats
2 has even multiplity
we just do 2 is odd and 1 is even so
for roots, r1 and r2, the facotrs would be
(x-r1)(x-r2)
so
(x-(-1))^1(x-2)^2(x-4)
(x+1)(x-2)^2(x-4)
this is a 4th degre equaton
normally, it is goig from top right to top left
it is upside down
theefor it has negative leading coefient



y=-k(x+1)(x-4)(x-2)^2

Answer 2

Solution-

General 4th degree polynomial equation with roots as a, b, c, d is,

$y=(x-a)(x-b)(x-c)(x-d)$

If the graph touches or bounces on a root, then that root has an even multiplicity.

If the graph goes through a root, then that root has an odd multiplicity.

Here, the roots are -1, 2, 4. (∵ as the graph crosses or touches x-axis or y=0 line)

At x=2, the graph touches the x-axis, so it has a multiplicity of 2

Now, we have to consider the end behavior or leading co-efficient.$As\ x\rightarrow -\infty,\ f(x)\rightarrow -\infty$

and

$As\ x\rightarrow +\infty,\ f(x)\rightarrow -\infty$

So, the graph must be function with even degree and negative leading co-efficient.

So, the final equation becomes,

$y=-(x+1)(x-2)^2(x-4)$