How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry of #y=-4(x-2)^2+2#?

2 Answers
Apr 26, 2017

Given the vertex form #y = a(x-h)^2+k" [1]"#

If #a >0#, then the graph opens up
If #a<0#, then the graph opens down.
The vertex is the point #(h,k)# and the axis of symmetry is #x = h#

Explanation:

For the given equation, #y=-4(x-2)^2+2#, we observe that #a = -4, h = 2 and k =2#

-4 < 0, therefore, the graph opens down.

The vertex is the point #(2,2)#

The axis of symmetry is #x = 2#

Apr 26, 2017

#"down " (2,2),x=2#

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 constant.

#• "if " a>0" then graph opens up " uuu#

#"if " a<0" then graph opens down " nnn#

#"here " a=-4rArr" graph opens down"#

#y=-4(x-2)^2+2" is in vertex form with "#

#h=2" and " k=2#

#rArrcolor(magenta)"vertex "=(h,k)=(2,2)#

The axis of symmetry is vertical and passes through the vertex

#rArr" equation of axis of symmetry is " =2#
graph{(y+4x^2-16x+14)(y-1000x+2000)=0 [-10, 10, -5, 5]}