possible equation

[red and white kop!]

im supposed to find a possible equation for this graph; i know that the equation must be degree 6 cos it has five turning points but otherwise i have on idea where to start, all of my attempts on graphic calculator failed miserably.
any hints?

 

[soroban]

Hello, red and white kop!


I'll assume the graph symmetric to the $\displaystyle y$-axis.

Choose six $\displaystyle x$-intercepts, say: .$\displaystyle \text{-}3, \text{-}2, \text{-}1, 1, 2, 3$

Then the function is of the form: .$\displaystyle f(x) \:=\:a(x+3)(x+2)(x+1)(x-1)(x-2)(x-3)$

Since the graph "opens downward", $\displaystyle a$ is negative.


One possible function is: .$\displaystyle f(x) \;=\;-(x+3)(x+3)(x+1)(x-1)(x-2)(x-3)$

. . . which simplifies to: .$\displaystyle f(x) \;=\;-x^6 + 14x^4 - 49x^2 + 36$

 

[red and white kop!]

but doesnt this make the y-values way too high?

 

[soroban]

Hello again, red and white kop!

but doesnt this make the y-values way too high?

There is no scale on your diagram.
. . How can the y-values be "too high"?


$\displaystyle \text{My graph has: }\;\begin{Bmatrix}\text{Maxima:} & \left(\pm\sqrt{7},36\right) \\ & (0,36) \\ \\ \text{Minima:} & \left(\pm\sqrt{\frac{7}{3}}, \text{-}\frac{400}{27}\right) & \approx & (\pm1.53, -14.81) \end{Bmatrix}$

 

[red and white kop!]

i guess you're right, but still is there any way to decrease the y-values without tinkering with the x-intercepts?
i guess not

 

[galactus]

If you want to decrease the y values but leave the x intercepts alone, simply change the coefficient. Notice the 'a' Soroban has in his first post. Make that something besides -1 to change the heught of the y's


$\displaystyle \;\ \text{change this value. Soroban used -1}$
$\displaystyle \downarrow$
$\displaystyle \fbox{a}(x+3)(x+2)(x+1)(x-1)(x-2)(x-3)$