how to find vertex for an x to the fourth??

i have an equation like

3x^4-8x^3+9x^2+22x-6 (i changed coefficents)

how do i graph this? like in general, because i dont know if the equation above is real or not.. i just didnt want to post the real equation.

my main question is how to find the vertex. i know -b/2a, but whats a and b and what would be c?

Top 2 Answers

A few days ago

Anonymous

Favorite Answer

You won’t be able to use an easy formula like the Quadratic Formula to solve a quartic like this. There is no such formula. Similarly, there really isn’t a “vertex” per se for a quartic. There are techniques for finding the x-intercepts of such a function, and calculus gives you techniques for finding local maxima and minima (peaks and valleys, if you will), but honestly there’s no way I can teach you that in a forum like this. You might try googling “Descartes’ Rule of Signs” and the “Rational Root Theorem” as a good place to start.

I can tell you this: the function you’ve written crosses the x-axis exactly once when x < 0 and either one or three times when x > 0. (Descartes’ Rule of Signs – very easy to apply.)

4 years ago

Use graph paper and discover the three given factors. To get from Q to P, you bypass down 2, left 5. To get from R (thirteen,4) to the fourth vertex, do an analogous, down 2, left 5. you’re at (8,5) you’re employing the certainty that the slopes are 2/5 you are able to tutor that the different 2 are = – 5/2 to tutor that this is rectangle.

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help im stuck?

how to find vertex for an x to the fourth??

how to find vertex for an x to the fourth??

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