How do you graph the parabola #f(x) = -(x+5)^2# using vertex, intercepts and additional points?

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Answer 1 · 2018-02-25T00:37:43

See below:

This is already in vertex form which is

#y=a(x-h)^2+k# where #(h,k)# is the vertex,

Here the vertex form can be rewritten as

#y=-1(x+5)^2+0# with #(-5,0)# as the vertex.

To find the y-intercept, plug in #0# for #x#:

#y=-1(0+5)^2+0#

#y=-1*25#

#y=-25#

The y-intercept is at #(0,-25)#

To find the x-intercept, plug in #0# for #y#:

#0=-1(x+5)^2+0#

#0=-1(x^2+10x+25)#

#0=-x^2-10x-25#

Factor:

#0= -1(x^2+10x+25)#

#0=-1(x+5)^2#

As you can see, you get the vertex form again, so you didn't have to expand.

The x-intercept is at #(5,0)#, which is also the vertex.

To find additional points, plug in values for #x#:

#y=-1(1+5)^2=-36->(1,-36)#

#y=-1(2+5)^2=-49->(2,-49)#

#y=-1(-1+5)^2=-16->(-1,-16)#

#y=-1(-2+5)^2=-9->(-2,-9)#

Here is a graph:

graph{-(x+5)^2 [-14, 6, -8.08, 1.92]}