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?
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).
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)
To find the possible values of d for the quadratic function f(x) = x^2 + dx + 3d to have its vertex on the x-axis, we need to consider the vertex form of a quadratic function.
In vertex form, a quadratic function is expressed as f(x) = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex.
For the given function f(x) = x^2 + dx + 3d, we can rewrite it in vertex form by completing the square:
f(x) = x^2 + dx + 3d
= (x^2 + dx) + 3d
= (x^2 + dx + d^2) - d^2 + 3d
= (x + d/2)^2 - d^2 + 3d
Comparing this to the vertex form, we can see that the vertex of the function occurs when x + d/2 = 0, or x = -d/2. Since the vertex is on the x-axis, the y-coordinate of the vertex (which is given by the expression -d^2 + 3d) must be zero.
-d^2 + 3d = 0
d(3 - d) = 0
This equation has two possible solutions, d = 0 and d = 3. Therefore, the possible values of d for the function f(x) = x^2 + dx + 3d to have its vertex on the x-axis are d = 0 and d = 3.
For the second quadratic function f(x) = x^2 + 3dx − d^2 + 1, we can use the same process to find the possible values of d.
f(x) = x^2 + 3dx − d^2 + 1
= (x + 3d/2)^2 - (d^2 - 1)
To have the vertex on the x-axis, the y-coordinate (-d^2 + 1) must be zero:
-d^2 + 1 = 0
d^2 = 1
d = ±1
Therefore, the possible values of d for the function f(x) = x^2 + 3dx − d^2 + 1 to have its vertex on the x-axis are d = 1 and d = -1.
To find the possible values of d for a quadratic function whose vertex is on the x-axis, we need to determine when the y-coordinate (or output) of the vertex is zero.
In general, the x-coordinate of the vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by x = -b / (2a). Since the vertex lies on the x-axis, this means that the y-coordinate is zero.
Let's apply this to the given quadratic function f(x) = x^2 + dx + 3d:
1. Set f(x) = 0 and solve for x:
x^2 + dx + 3d = 0
2. Use the formula for the x-coordinate of the vertex:
x = -d / (2*1)
3. Set the y-coordinate equal to zero and solve for d:
0 = (-d / (2*1))^2 + d * (-d / (2*1)) + 3d
Simplifying this equation will give us the possible values of d.
Now, let's consider the quadratic function f(x) = x^2 + 3dx - d^2 + 1:
1. Set f(x) = 0 and solve for x:
x^2 + 3dx - d^2 + 1 = 0
2. Use the formula for the x-coordinate of the vertex:
x = -3d / (2*1)
3. Set the y-coordinate equal to zero and solve for d:
0 = (-3d / (2*1))^2 + 3d * (-3d / (2*1)) - d^2 + 1
Again, simplifying this equation will give us the possible values of d.
To find the exact values of d, the equations need to be solved algebraically. The resulting equations may be factorable or require the use of the quadratic formula.