What is the vertex of the graph of #y = 2(x – 3)^2 + 4#?

2 Answers
Shwetank Mauria
Jul 2, 2018

Vertex is #(3,4)#

Explanation:

If the equation of parabola is of the form #y=a(x-h)^2+k#,

the vertex is #(h,k)#.

Observe that when #x=h#, the value of #y# is #k# and as #x# moves on either side, we have #(x-h)^2>0# and #y# rises.

Hence, we have a minima at #(h,k)#. It would be maxima if #a<0#

Here we have #y=2(x-3)^2+4#, hence we have vertex at #(3,4)#, where we have a minima.

graph{2(x-3)^2+4 [-6.58, 13.42, 0, 10]}

Jim G.
Jul 2, 2018

#"vertex "=(3,4)#

Explanation:

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

#•color(white)(x)y=a(x-h)^2+k#

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

#y=2(x-3)^2+4" is in this form"#

#"with "(h,k)=(3,4)larrcolor(magenta)"vertex"#

#"and "a=2#

#"since "a>0" then graph is a minimum"#
graph{2(x-3)^2+4 [-20, 20, -10, 10]}

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