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

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Answer 1 · 2017-02-27T21:34:31

Maximum $\to (x, y) = (\frac{3}{4}, \frac{27}{8})$ $->(x,y)=(3/4,27/8)$

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

The maximum is ta the vertex.

Write as: $y = - 6 (x^{2} - \frac{9}{6} x)$ $y=-6(x^2-9/6x)$

$x_{vertex} = (- \frac{1}{2}) \times (- \frac{9}{6}) = + \frac{9}{12} = \frac{3}{4}$ $x_("vertex")=(-1/2)xx(-9/6) = + 9/12= 3/4$

By substitution we have:

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

$y_{vertex} = + \frac{27}{8}$ $y_("vertex")=+27/8$

Thus the maximum $\to (x, y) = (\frac{3}{4}, \frac{27}{8})$ $->(x,y)=(3/4,27/8)$

Tony B Tony B

How do you find the max or minimum of f(x)=−6x2+9xf(x)=-6x^2+9x?