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What real value of t produces the smallest value of the quadratic t^2-9t-36?

Jul 13, 2021

#1
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Minimizing, to obtain t^2-9t-36 = -225/4. The value of t is 9/2

Jul 13, 2021
#2
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What real value of t produces the smallest value of the quadratic t^2-9t-36?

t2 – 9t – 36 factors to (t – 12)(t + 3).   The smaller value there would be t = –3

I read the previous answer first, and I can't figure out where the –225/4 came from.

Jul 13, 2021
#4
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Hi xy Thanks for answering HSDx and guest But, guest, my answer is for you to learn from as well What real value of t produces the smallest value of the quadratic t^2-9t-36?

maybe it will make more sense to you change the t to an x and then put it equal to y

\(y=x^2-9x-36 \)

Now you are being asked for the smallest value of x

If you graph this you will get a concave up parabola.

I know this becasue the x^2 makes it a parabola

and the invisable number in front of the x^2 is  +1.

Since it is + the parabola will be concave up.

The lowest point of a concave up parabola will be the turning point. The x value for it which will be halfway between the roots.

roots are when y=0 so

\(0=x^2-9x-36\\ 0=(x-12)(x+3)\\ x=12\;\;or \;\;x=-3 \)

Halfway between the roots is  x= (12+-3)/2 = 4.5

That is the x value (actually the t value) that you want.     t=4.5

that is the value that will give the expression the smallest value.

If you wanted to find that smallest value then you would sub 4=4.5 back into the expression. Melody  Jul 13, 2021
edited by Melody  Jul 13, 2021
#5
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Melody, thank you for going to the trouble to explain that to me.  I get it.  Your explanation made it easy to understand.

Guest Jul 14, 2021
#7
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Thanks very much guest, I am glad I could help Melody  Jul 14, 2021