Writing a Quadratic Function in Factored Form with Given Vertex and Two Distinct Roots: Key Concepts and Step-By-Step Guide

Explain how you could write a quadratic function in factored form that would have a vertex with an x-coordinate of 3 and two distinct roots.

sample response : The vertex is on the axis of symmetry, so the axis of symmetry is x = 3. Find any two x-intercepts that have the equivalent distance from the axis of symmetry. Use those x-intercepts to write factors of the function by subtracting their values from x. For example, 2 and 4 are each 1 unit from x = 3, so f(x) = (x – 2)(x – 4) is a possible function.

To write a quadratic function in factored form with a given vertex and two distinct roots, you need to understand a few key concepts.

1. The vertex form of a quadratic function is given by:
f(x) = a(x – h)^2 + k
Here, (h, k) represents the coordinates of the vertex.

2. Factored form of a quadratic function can be expressed as:
f(x) = a(x – r1)(x – r2)
where r1 and r2 are the roots of the quadratic equation.

Given that the x-coordinate of the vertex is 3, we can say that h = 3. Now, we need to find the values of r1 and r2. Since we want two distinct roots, we can set the equation equal to zero and solve for x.

Let’s consider a generic quadratic equation:
f(x) = a(x – r1)(x – r2)

When factored, this equation becomes:
f(x) = ax^2 – a(r1 + r2)x + ar1r2

We can equate this to zero to find the roots:
ax^2 – a(r1 + r2)x + ar1r2 = 0

Since we want the x-coordinate of the vertex to be 3, we know that (-b/2a) = 3, where b = a(r1 + r2).
Simplifying this equation, we get:
-2(r1 + r2) = 6
r1 + r2 = -3 [Equation 1]

To find two distinct roots, r1 ≠ r2, and the product of the roots is given by ar1r2.

Now, let’s consider the quadratic function:
f(x) = a(x – r1)(x – r2)

Since we want distinct roots, we can assume r1 = 3 – p and r2 = 3 + p,
where p ≠ 0 represents the difference between the two roots.

Based on Equation 1, we know that:
r1 + r2 = -3
(3 – p) + (3 + p) = -3
6 = -3 (which is not true)

Hence, the assumption r1 = 3 – p and r2 = 3 + p does not hold, and we need to reassess. To have distinct roots, we cannot use this assumption.

However, if we allow for the possibility of complex roots, we can proceed.

Let’s assume that the quadratic function is:
f(x) = a(x – (3 – ip))(x – (3 + ip))
= a[(x – 3 + ip)(x – 3 – ip)]

Expanding this equation, we get:
f(x) = a(x^2 – 3x + ipx – 3x + 9 – 3ipx – ipx + 3ip – ipx + 3ip + i^2p^2)
= a(x^2 – 6x + 9 – i^2p^2)
= a(x^2 – 6x + 9 + p^2)
= ax^2 – 6ax + 9a + ap^2

Now, we need to determine the value of a and p.

Since the x-coordinate of the vertex is 3, we can identify that the axis of symmetry is x = 3. Therefore, the value of x at the vertex is equal to 3. Plugging in these values in our function, we get:

f(3) = a(3^2) – 6a(3) + 9a + ap^2
= 9a – 18a + 9a + ap^2
= 0 + ap^2
= ap^2

Since the vertex has an x-coordinate of 3, we can say that p = ±√3.

Hence, our final quadratic function in factored form with the given conditions is:
f(x) = ax^2 – 6ax + 9a ± a√3^2
= ax^2 – 6ax + 9a ± 3a

This function has a vertex with an x-coordinate of 3 and two distinct roots, considering that we allow for the possibility of complex roots.

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