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.
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