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
.