How do you find the coordinates of the vertex $y = - 6 x^{2} + 14 x$ $y= -6x^2 + 14x$ ?

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Answer 1 · 2016-08-13T11:15:08

$P = (\frac{7}{6}, 10.5)$ $P=(7/6,10.5)$

$y = - 6 x^{2} + 14 x$ $y=-6x^2+14x$

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$Let the coordinates of vertex be P (x, y)$ $"Let the coordinates of vertex be "P(x,y)$

$Take derivative of the function$ $"Take derivative of the function "$

$\frac{d y}{d x} = - 12 x + 14$ $(d y)/(d x)=-12x+14$

$Now calculate \frac{d y}{d x} = 0$ $"Now calculate "(d y)/(d x)=0$

$- 12 x + 14 = 0$ $-12x+14=0$

$12 x = 14$ $12x=14$

$x = \frac{14}{12} = \frac{7}{6} (This is the x coordinate of vertex)$ $x=14/12=7/6 " (This is the x coordinate of vertex)"$

$calculate f (x) to find the y coordinate of vertex$ $"calculate "f(x) " to find the y coordinate of vertex"$

$plug x= \frac{7}{6} into y$ $"plug x=" 7/6 " into y"$

$y = - 6 \cdot {(\frac{7}{6})}^{2} + 14 \cdot \frac{7}{6}$ $y=-6*(7/6)^2+14*7/6$

$y = - \frac{49}{6} + \frac{98}{6}$ $y=-49/6+98/6$

$y = \frac{49}{6}$ $y=49/6$

$y = (This is the y coordinate of vertex)$ $y=("This is the y coordinate of vertex")$

$Thus$ $"Thus"$

$P = (\frac{7}{6}, 8.17)$ $P=(7/6,8.17)$

How do you find the coordinates of the vertex y=−6x2+14xy= -6x^2 + 14x?