What is the vertex of $y = \frac{3}{2} x^{2} + 20 x + 21$ $y = 3/2x^2+20x+21$ ?

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What is the vertex of $y = \frac{3}{2} x^{2} + 20 x + 21$ $y = 3/2x^2+20x+21$ ?

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Answer 1 · 2018-06-23T13:30:28

$vertex = (- \frac{20}{3}, - \frac{137}{3})$ $"vertex "=(-20/3,-137/3)$

Explanation:

$given a parabola in standard form$ $"given a parabola in "color(blue)"standard form"$

$∙ x y = a x^{2} + b x + c x; a \neq 0$ $•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0$

$then the x-coordinate of the vertex is$ $"then the x-coordinate of the vertex is "$

$∙ x x_{vertex} = - \frac{b}{2 a}$ $•color(white)(x)x_(color(red)"vertex")=-b/(2a)$

$y = \frac{3}{2} x^{2} + 20 x + 21 is in standard form$ $y=3/2x^2+20x+21" is in standard form"$

$with a = \frac{3}{2}, b = 20 and c = 21$ $"with "a=3/2,b=20" and "c=21$

$x_{vertex} = - \frac{20}{3}$ $x_("vertex")=-20/3$

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

$y_{vertex} = \frac{3}{2} {(- \frac{20}{3})}^{2} + 20 (- \frac{20}{3}) + 21$ $y_("vertex")=3/2(-20/3)^2+20(-20/3)+21$

$\times \times = \frac{200}{3} - \frac{400}{3} + \frac{63}{3} = - \frac{137}{3}$ $color(white)(xxxx)=200/3-400/3+63/3=-137/3$

$vertex = (- \frac{20}{3}, - \frac{137}{3})$ $color(magenta)"vertex "=(-20/3,-137/3)$

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What is the vertex of $y = \frac{3}{2} x^{2} + 20 x + 21$ $y = 3/2x^2+20x+21$ ?

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Explanation:

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