How do you find the max or minimum of $f(x)=-6x^2+9x$ ?

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Answer 1 · 2017-02-27T21:34:31

Maximum $->(x,y)=(3/4,27/8)$

The coefficient of $x^2$ is negative so the graph is of form $nn$ thus the vertex is a maximum.

The maximum is ta the vertex.

Write as: $y=-6(x^2-9/6x)$

$x_("vertex")=(-1/2)xx(-9/6) = + 9/12= 3/4$

By substitution we have:

$y_("vertex")=-6(3/4)^2+9(3/4)$

$y_("vertex")=+27/8$

Thus the maximum $->(x,y)=(3/4,27/8)$

Tony B Tony B

How do you find the max or minimum of f(x)=-6x^2+9xf(x)=-6x^2+9x?