How do you graph #y=-2(3+x)^2 +4#?

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Answer 1 · 2015-08-06T23:01:32

Calculate the vertex and the #x#- and #y#-intercepts and then sketch the graph.

#y=-2(3+x)^2+4#

Step 1. Convert your equation to vertex form.

The vertex form is #y = a(x-h)^2 +k#.

Rearrange your equation to #y = -2(x+3)^2 +4#.

Then #a=-2#, #h=-3#, and #k=4#.

Step 2. Find the vertex.

The vertex is at (#h,k#) or (#-3,4#).

Step 3. Find the #y#-intercept.

Set #x=0# and solve for #y#.

#y=-2(3+x)^2+4 = -2(3)^2+4 = -2(9) +4 = -18+4 = -14#

The #y#-intercept is at (#0,-14#).

Step 4. Find the #x#-intercept(s).

Set #y=0# and solve for #x#.

#0 = -2(3+x)^2+4#

#(3+x)^2 -2 =0#

#((3+x)+sqrt2)( (3+x)-sqrt2) = 0#

#3+x+sqrt2=0# and #3+x-sqrt2=0#

#x=-3-sqrt2≈ -4.4# and #x=-3+sqrt2≈-1.6#

Step 5. Draw your axes and plot the four points.

Graph 1

Step 6. Draw a smooth parabola that passes through the four points.

Graph 2

And you have your graph.