Ask your own question, for FREE!
Mathematics 79 Online
Subshilava:

Help Create a function that meets the following requirements. Has 3 turning points Has the following zeros: -2, 4 (Note: Your function can have more zeros than these)

Luigi0210:

Do you at least know how to start it off?

Subshilava:

I know that the zeros would be (x+2) and (x-4) and that since it had three turning points I need another zero

Luigi0210:

Yes, that's correct, you at least have \\[\large (x+2)(x-4)=0 \] However, this will only produce an \( x^2\) function, which has ONE turning point. Do you know any way to add more turning points?

Subshilava:

Expand it?

Luigi0210:

Even though it's not explicitly stated, zero could be a zero of the function.

Subshilava:

Srry my moms texting me 😪

Luigi0210:

In simpler terms, if you add a zero, it adds a turning point: \(\large x(x+2)(x-4)=0 \) gives you two turning points. If you add another 0 \(\large x^2(x+2)(x-4)=0 \) , that gives you three turning points. You can graph to confirm that it has three turning points.

Subshilava:

Oh ok I forgot about 0’s ty

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!