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

if the vertex is on the x-axis, say, at (h,0) then we know that
y = a(x-h)^2
So, we know that
a(x^2-2hx+h^2) = x^2+dx+3d
So we know a=1, and thus
x^2-2hx+h^2 = x^2+ dx+3d
-2h = d
h^2 = 3d = -6h
So, h = -6 and we have d = 12
f(x) = x^2 + 12x + 36 = (x+6)^2
This has its vertex at (-6,0)

Now go through the same logic for the other f(x).

Answer 2

or
complete the square:
f(x) = x^2 + dx + 3d
= x^2 + dx + d^2/4 - d^2/4 + 3d
= (x + d/2)^2 + 3d-d^2/4
so the vertex is (-d/2 , 3d - d^2/4)
but on the x-axis, the y value must be zero
3d - d^2/4 = 0
12d - d^2 = 0
d(12 - d) = 0
so d = 0 or d = 12

the equation could have been f(x) = x^2 with vertex (0,0)
or f(x) = x^2 + 12x + 36, with vertex (-6,0)