Quadratic

 

What real value of $t$ produces the smallest value of the quadratic $t^2 -9t - 36 + 13t - 60$?      

 

Combine like terms and arrange in standard  

quadratic format like ax² + bx + c                         t² + 4t – 96    

 

The t² tells us the curve is a parabola, and the fact   

 that t² is positive tells us the parabola opens upward   

 

The first derivitive of the equation describes the slope of the curve  

The vertex is the place where the curve turns around and goes back up,  

therefore is the smallest value, and the slope is zero at the vertex       

 

The first deritive is                                               2t + 4  

 

Set it equal to zero and solve for t                       2t + 4 = 0   

                                                                                    t = –2   

This is the answer to "What real value of t" 

but let's keep going and find out just what   

that smallest value actually is   

 

Substitute –2 for t back into original equation        (–2)² + 4(–2) – 96    

                                                                                   4 – 8 – 96 = –100   

 

So the smallest value is –100 which occurs when t = –2    

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