Precalc question

Question

If the graph of the quadratic function f(x) = x^2 + dx + 3d has its vertex on the x-axis, what are the possible values of d? What if f(x) = x^2+3dx−d^2+1

Answer 1

f(x)=x^2+dx+3d has its vertex on the x-axis which means that in (h,k), the vertex, k=0

y=x^2+dx +3d

complete the square

(d/2)^2=d^2/4

y=x^2+dx+d^2/4+3d-d^2/4

since k=0, 3d-d^2/4=0

3d-d^2/4=0

12d-d^2=0

d^2-12d=0

d(d-12)=0

d=0 or d=12

if d=0 you get y=x^2

if d=12 you get....

y=x^2+dx+3d

complete the square

y=x^2+12x +36

y=x^2+12x+36 +36-36

y=(x+6)^2

in either case, k=0

y=x^2+3dx-d^2+1

follow the same reasoning to get...

y=x^2+3dx -d^2+1

y=x^2+3dx+9d^2/4 -d^2+1-9d^2/4

again, k=0, so.....

-d^2+1-9d^2/4=0

simplify

(-13/4)d^2+1=0

(13/4)d^2=1

d^2=4/13

d=±√(4/13)

d=±√4/√13

d=±2/√13

d=±2√13/13

so d=2√13/13 or d=-2√13/13

check: -d^2+1-(9/4)d^2

-4/13+1-(9/4)(4/13)

-4/13-9/13+1=0