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What is the vertex of #7y= 12(x-15)^2 +12#?

2 Answers

Answer:

The vertex happens to be
#(x,y)=(15,12/7)#

Explanation:

The given equation is:
#7y=12(x-15)^2+12#
The curve is symmetrical about the x axis
Differentiating the equation wrt x
#7dy/dx=12(2)(x-15)+0#
The vertex coresponds to the point where the slope is zero.
Equating #dy/dx=0#
#7(0)=24(x-15)#
ie
#24(x-15)=0#
#x-15=0#
#x=15#
Substituting for x in the equation of the curve
#7y=12(15-15)+12#
#7y=12#
#y=12/7#
Thus, the vertex happens to be
#(x,y)=(15,12/7)#

Feb 9, 2018

Answer:

#"vertex "=(15,12/7)#

Explanation:

#"divide both sides by 7"#

#rArry=12/7(x-15)^2+12/7#

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#y=12/7(x-15)^2+12/7" is in vertex form"#

#rArrcolor(magenta)"vertex "=(15,12/7)#