How do you find the vertex of the parabola #y=-2x^2 + 12x - 13#?

2 Answers
Mar 15, 2018

#"vertex "=(3,5)#

Explanation:

#"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"#

#"to obtain this form use the method of "color(blue)"completing the square"#

#• " the coefficient of the "x^2" term must be 1"#

#rArry=-2(x^2-6x+13/2)#

#• " add/subtract "(1/2"coefficient of the x-term")^2" to"#
#x^2-6x#

#y=-2(x^2+2(-3)xcolor(red)(+9)color(red)(-9)+13/2)#

#color(white)(y)=-2(x-3)^2-2(-9+13/2)#

#color(white)(y)=-2(x-3)^2+5larrcolor(red)"in vertex form"#

#rArrcolor(magenta)"vertex "=(3,5)#
graph{(y+2x^2-12x+13)((x-3)^2+(y-5)^2-0.04)=0 [-10, 10, -5, 5]}

Mar 15, 2018

Vertex is at #(3,5)# .

Explanation:

#y = -2x^2+12x-13 or y = -2(x^2-6x)-13 # or

#y = -2(x^2-6x +9)+18-13 # or

#f(x) = -2(x-3)^2+5 #. Comparing with vertex form of

equation #y = a(x-h)^2+k ; (h,k)# being vertex we find

here #h=3 , k=5 :.# Vertex is at #(3,5)# .

graph{-2x^2+12x-13 [-17.78, 17.78, -8.89, 8.89]} [Ans]