+5 What real value of t produces the smallest value of the quadratic t^2-9t-36?
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.
+118609 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.
