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# Can anyone help me with this?

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What real value of t produces the smallest value of the quadratic t^2 -9t - 36?

Jul 26, 2019

#1
+2

To find the minimum value, you can complete the square.

Using the formula $$a(t+d)^2+e$$ where $$d=\frac{-b}{2a}$$ and $$e=c-\frac{b^2}{4a}$$, the completed square would be $$(t-\frac{9}{2})^2-45$$. Since squares cannot be negative, we see that the minimum value would be where $$(t-\frac{9}{2})^2$$ is 0. Thus $$t=\frac{9}{2}$$

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Jul 26, 2019
#2
+19832
+3

There are a couple of other ways....   this is a bowl shaped parabola (because the leading coefficient is positive)

- b / 2a

= 9/2

or take the derivative and = 0

2t-9 = 0

2t=9

t=9/2

Jul 26, 2019