What is the vertex of $y = - 3 x^{2} + 5 x + 6$ $y=-3x^2+5x+6$ ?

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What is the vertex of $y = - 3 x^{2} + 5 x + 6$ $y=-3x^2+5x+6$ ?

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Answer 1 · 2017-06-20T07:15:07

$0.833, 8.083$ $0.833, 8.083$

Explanation:

The vertex can be found using differentiation, differentiating the equation and solving for 0 can determine where the x point of the vertex lies.

$\frac{d y}{d x} (- 3 x^{2} + 5 x + 6) = - 6 x + 5$ $dy/dx (-3x^2 + 5x +6) = -6x + 5$
$- 6 x + 5 = 0, 6 x = 5, x = \frac{5}{6}$ $-6x + 5 = 0, 6x = 5, x = 5/6$

Thus the $x$ $x$ coordinate of the vertex is $\frac{5}{6}$ $5/6$
Now we can substitute $x = \frac{5}{6}$ $x = 5/6$ back into the original equation and solve for $y$ $y$ .

$y = - 3 {(\frac{5}{6})}^{2} + 5 (\frac{5}{6}) + 6$ $y = -3(5/6)^2 + 5(5/6) + 6$
$y = 8.0833$ $y = 8.0833$

Answer 2 · 2017-06-20T07:19:15

$(\frac{5}{6}, \frac{97}{12})$ $(5/6,97/12)$

Explanation:

$for a parabola in standard form y = a x^{2} + b x + c$ $"for a parabola in standard form " y=ax^2+bx+c$

$the x-coordinate of the vertex is x_{vertex} = - \frac{b}{2 a}$ $"the x-coordinate of the vertex is " x_(color(red)"vertex")=-b/(2a)$

$y = - 3 x^{2} + 5 x + 6 is in standard form$ $y=-3x^2+5x+6" is in standard form"$

$with a = - 3, b = 5, c = 6$ $"with " a=-3,b=5,c=6$

$\Rightarrow x_{vertex} = - \frac{5}{- 6} = \frac{5}{6}$ $rArrx_(color(red)"vertex")=-5/(-6)=5/6$

$substitute this value into the function for y-coordinate$ $"substitute this value into the function for y-coordinate"$

$\Rightarrow y_{vertex} = - 3 {(\frac{5}{6})}^{2} + 5 (\frac{5}{6}) + 6 = \frac{97}{12}$ $rArry_(color(red)"vertex")=-3(5/6)^2+5(5/6)+6=97/12$

$\Rightarrow vertex = (\frac{5}{6}, \frac{97}{12})$ $rArrcolor(magenta)"vertex "=(5/6,97/12)$

Answer 3 · 2017-06-20T07:22:20

$(\frac{5}{6}, \frac{97}{12})$ $(5/6,97/12)$

Explanation:

$y = a x^{2} + b x + c$ $y=ax^2+bx+c$ [Standard Form of a Quadratic Equation]
$y = - 3 x^{2} + 5 x + 6$ $y=-3x^2+5x+6$

$a = - 3$ $a = -3$
$b = 5$ $b = 5$
$c = 6$ $c = 6$

TO FIND THE X-VALUE OF THE VERTEX:
Use the formula for the axis of symmetry by substituting values for $b$ $b$ and $a$ $a$ :
$x = \frac{- b}{2 a}$ $x = (-b)/(2a)$
$x = \frac{- 5}{2 (- 3)}$ $x = (-5)/(2(-3))$
$x = \frac{- 5}{- 6}$ $x = (-5)/-6$
$x = \frac{5}{6}$ $x = 5/6$

TO FIND THE Y-VALUE OF THE VERTEX:
Use the formula below by substituting values for $a$ $a$ , $b$ $b$ , and $c$ $c$ :
$y = \frac{- b^{2}}{4 a} + c$ $y = (-b^2)/(4a)+c$
$y = \frac{- {(5)}^{2}}{4 (- 3)} + 6$ $y = (-(5)^2)/(4(-3))+6$
$y = \frac{- 25}{- 12} + 6$ $y = (-25)/(-12)+6$
$y = \frac{25}{12} + \frac{72}{12}$ $y = 25/12+72/12$
$y = \frac{97}{12}$ $y = 97/12$

Express as a coordinate.
$(\frac{5}{6}, \frac{97}{12})$ $(5/6,97/12)$

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