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

3 Answers
Jun 20, 2017

#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.

#dy/dx (-3x^2 + 5x +6) = -6x + 5#
#-6x + 5 = 0, 6x = 5, x = 5/6#

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

#y = -3(5/6)^2 + 5(5/6) + 6#
#y = 8.0833#

Jun 20, 2017

#(5/6,97/12)#

Explanation:

#"for a parabola in standard form " y=ax^2+bx+c#

#"the x-coordinate of the vertex is " x_(color(red)"vertex")=-b/(2a)#

#y=-3x^2+5x+6" is in standard form"#

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

#rArrx_(color(red)"vertex")=-5/(-6)=5/6#

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

#rArry_(color(red)"vertex")=-3(5/6)^2+5(5/6)+6=97/12#

#rArrcolor(magenta)"vertex "=(5/6,97/12)#

Jun 20, 2017

#(5/6,97/12)#

Explanation:

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

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

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

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

Express as a coordinate.
#(5/6,97/12)#