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

Jan 17, 2024

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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

.

Jan 17, 2024