If the graph of the quadratic function fx x2 dx 4d has it

If the graph of the quadratic function f(x) = x^2 + dx + 4d has its vertex on the x-axis, what are the possible values of d? (Enter your answers as a comma-separated list.) d = -x^2/x 4 what if f(x) = x^2 + 3dx -d^2 + 1? (Enter your answers as a comma-separated list.) d = -3x/2 + Squareroot 13x^2+4/2, -3x/2- Squareroot13x^2+4/2

Solution

The given quadratic function is y = f(x) = x2+dx +4d = (x2+2*x*d/2 + d2/4) +(-d2/4+ 4d) = (x+d/2)2+ (4d- d2/4). This is the equation of a parabola, wth vertex at (-d/2, 4d- d2/4). Since the vertex is on the X-Axis, we must have 4d- d2/4 = 0 or, d(4-d/4) = 0. Thus, either d = 0 or, 4-d/4= 0, in which case, d = 16. Thus, the possible values of d are 0 and 16. If the given quadratic function is y = f(x) = x2+3dx –d2+1, then f(x) = (x+2*3d/2+ 9d2/4) –d2(1+9/4) +1 = (x+3d/2)2 +(1–13d2/4). ). This is the equation of a parabola, wth vertex at (-3d/2, 1–13d2/4). Since the vertex is on the X-Axis, we must have 1–13d2/4 = 0 or, 13d2/4 = 1 or, d2 = 4/13. Then   d = ± 2/13.