# Math (Pre-Calculus)

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9t^4 + 6t^2 + 2

the problem is asking for the value of t at which the graph is at it's minimum.

what i did was set t^4 to x^2 and t^2 to x. So that means I have 9x^2 + 6x + 2.

Completing the square and the vertex formula gave me the same answer of -1/3 as x. So setting that equal to t^2 and solving gives me a no solution answer. The same thing happens when I try to solve it using the quad formula. 36-72 shows up in the radical, which is negative.

So I'm at a loss, my answer comes out to be undefined.

But the answer to the problem (as stated in the back of the book) is x=0. Also when I put it in a graphing calculator the minimum is indeed 2 at t=0.

As far as I know, undefined is not the same as 0. I must be doing something wrong. Any help is appreciated. Thank you.

• Math (Pre-Calculus) - ,

If it is precalculus, you're not expected to use derivatives, which incidentally will give you x=0 as the answer.

I do not know if you teacher has shown you a systematic method to calculate the minimum for a quartic. In this particular situation, we can find the answer rapidly by taking advantage of the way the function is made up.

f(x) = 9x⁴ + 6x² + 2
is made up of three terms, two of which are non-negative. So the minimum is 2 when both of the non-negative terms vanish. It is obvious that 9x⁴ and 6x² vanish at x=0. The vertex is therefore (0,2).

I also note that f(x) can be composed by goh(x) where g(x)=9x²+6x+2 and h(x)=x². Not that this property will help us find the vertex without the use of calculus, as far as I know.

Also, I have not come across a method to find the vertex for a general quartic without the use of calculus. If anyone would share the answer, it would be appreciated.

• Math (Pre-Calculus) -corr. - ,

last paragraph: `"the vertices for a general quartic ".

Also, I have not mentioned that graphing by tabulation almost always works for finding extrema of polynomials, although a good judgment goes a long way.