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What real value of t produces the smallest value of the quadratic t^2 -9t - 36 + 4 - 30t?

 Dec 19, 2023
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What real value of t produces the smallest value of the quadratic t^2 -9t - 36 + 4 - 30t?  

 

Combine like terms                                    t2 – 39t – 32   

 

This is a parabola that opens upward.  

 

The smallest value will occur at the vertex.  That's the point where the curve turns around and heads back up. 

The first derivitive expresses the slope.  At the vertex, the slope equals zero.  Setting the slope equal to y:   

 

                                                                y  =  t2 – 39t – 32   

 

So the first derivitive is                             y'  =  2t – 39  

 

Set equal to zero                                      2t – 39  =  0 

 

                                                                   t  =  19.5    

 

The problem doesn't ask what   

the smallest value is, but you  

can find it by plugging 19.5  

into the original equation                    (19.5)2 – 39(19.5) – 32  

 

                                                            380.25 – 760.5 – 32  =  – 412.25   

 

Checked answer by pasting y=t^2–39t–32 into Desmos.  

 

After re-reading answer, I edited one word for terminology correction.  

.

 Dec 19, 2023
edited by Bosco  Dec 19, 2023

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