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

eramsby1O1O May 24, 2024

#1**0 **

First off, let's simplify and combine all like terms.

We get \(t^2+4t-96\). Now, we have a perfect parabola we can graph! The vetex of the graph is the minimum value point, and taking the x value from that coordinate pair will get us t.

The vertex we get is (-2, -100), which is our minimum point.

So t=-2 is our answer!

Thanks! :)

NotThatSmart May 24, 2024

#2**+2 **

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

Combine like terms t^{2} + 4t – 96

Arbitrarily say this equals "y" y = t^{2} + 4t – 96

The "degree" of an equation is determined by the highest power of a variable in the equation.

Since the highest power is the squared term, we know the curve is a parabola that opens upward.

The smallest value of "y" is the bottom where it turns around and goes back up.

We can find this by taking the first derivitive of the equation and setting it equal to zero.

The first derivitive defines the slope of a curve at any point.

So the first deritive of t^{2} + 4t – 96 is **2t + 4**

Set equal to zero is 2t + 4 = 0

2t = –4

t = **–2**

I took the equation to www.desmos.com/calculator to verify the answer.

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Bosco May 24, 2024