Vertex from Standard Form
y=x^2+5x+6y=x2+5x+6 is the standard form for a quadratic equation, ax^2+bx+6ax2+bx+6, where a=1a=1, b=5b=5, and c=6c=6.
The vertex form is a(x-h)^2+ka(x−h)2+k, and the vertex is (h,k)(h,k).
In the standard form, h=(-b)/(2a)h=−b2a, and k=f(h)k=f(h).
Solve for hh and kk.
h=(-5)/(2*1)h=−52⋅1
h=-5/2h=−52
Now plug in -5/2−52 for xx in the standard form to find kk.
f(h)=k=(-5/2)^2+(5xx-5/2)+6f(h)=k=(−52)2+(5×−52)+6
Solve.
f(h)=k=25/4-25/2+6f(h)=k=254−252+6
The LCD is 4.
Multiply each fraction by an equivalent fraction to make all of the denominators 44. Reminder: 6=6/16=61
f(h)=k=25/4-(25/2xx2/2)+(6/1xx4/4)f(h)=k=254−(252×22)+(61×44)
Simplify.
f(h)=k=25/4-50/4+24/4f(h)=k=254−504+244
Simplify.
f(h)=k=-1/4f(h)=k=−14
Vertex (-5/2,-1/2)(−52,−12)
Vertex form: a(x-h)^2+ka(x−h)2+k
1(x+5/2)^2-1/41(x+52)2−14
(x+5/2)^2-1/4(x+52)2−14
