How do you find the vertex of $y = 2 x^{2} + 10 x + 8$ $y=2x^2+10x+8$ ?

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How do you find the vertex of $y = 2 x^{2} + 10 x + 8$ $y=2x^2+10x+8$ ?

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Answer 1 · 2018-03-30T07:36:53

$(- \frac{5}{2}, - \frac{9}{2})$ $(-5/2,-9/2)$

Explanation:

$Given the equation of a parabola in standard form$ $"Given the equation of a parabola in "color(blue)"standard form"$

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

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

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

$y = 2 x^{2} + 10 x + 8 is in standard form$ $y=2x^2+10x+8" is in standard form"$

$with a = 2, b = 10 and c = 8$ $"with "a=2,b=10" and "c=8$

$\Rightarrow x_{vertex} = - \frac{10}{4} = - \frac{5}{2}$ $rArrx_("vertex")=-10/4=-5/2$

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

$y_{vertex} = 2 {(- \frac{5}{2})}^{2} + 10 (- \frac{5}{2}) + 8 = - \frac{9}{2}$ $y_("vertex")=2(-5/2)^2+10(-5/2)+8=-9/2$

$\Rightarrow vertex = (- \frac{5}{2}, - \frac{9}{2})$ $rArrcolor(magenta)"vertex "=(-5/2,-9/2)$
graph{2x^2+10x+8 [-10, 10, -5, 5]}

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How do you find the vertex of $y = 2 x^{2} + 10 x + 8$ $y=2x^2+10x+8$ ?

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