What is the vertex of # y = 3/2x^2+20x+21 #?

1 Answer
Jim G.
Jun 23, 2018

#"vertex "=(-20/3,-137/3)#

Explanation:

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

#•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0#

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

#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#

#y=3/2x^2+20x+21" is in standard form"#

#"with "a=3/2,b=20" and "c=21#

#x_("vertex")=-20/3#

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

#y_("vertex")=3/2(-20/3)^2+20(-20/3)+21#

#color(white)(xxxx)=200/3-400/3+63/3=-137/3#

#color(magenta)"vertex "=(-20/3,-137/3)#

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