How do you write #f(x)=-6x^2-5x+1# in vertex form?

Question

2698 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-24T08:21:09

#y=-6(x+5/12)^2+49/24#

Vertex form is #f(x)=a(x-h)^2+k#

We have #f(x)=-6x^2-5x+1#

= #-6(x^2+5/6x)+1#

= #-6(x^2+2xx5/12xx x+(5/12)^2)-(-6)xx(5/12)^2+1#

= #-6(x+5/12)^2+25/24+1#

= #-6(x+5/12)^2+49/24#

and vertex is #(-5/12,-49/24)#

graph{-6x^2-5x+1 [-3.8, 3, -1, 2.4]}