How to Write a Quadratic Function in Factored Form with a Vertex at x = 3 and Two Distinct Roots

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.

To write a quadratic function in factored form with a vertex at x = 3 and two distinct roots, we need to consider the factored form of a quadratic equation:

f(x) = a(x – r1)(x – r2)

In this form, r1 and r2 represent the roots of the equation, and a is a coefficient that determines the shape and scale of the parabola

To write a quadratic function in factored form with a vertex at x = 3 and two distinct roots, we need to consider the factored form of a quadratic equation:

f(x) = a(x – r1)(x – r2)

In this form, r1 and r2 represent the roots of the equation, and a is a coefficient that determines the shape and scale of the parabola.

Given that the vertex has an x-coordinate of 3, we know the vertex form of the equation can be expressed as:

f(x) = a(x – 3)(x – r2)

To ensure the quadratic has two distinct roots, we need to set r1 and r2 as different values. Let’s assume r1 is the root that corresponds to the x-coordinate of the vertex (r1 = 3), and r2 is the other distinct root. So, let’s introduce r2 as a variable, and rewrite the equation:

f(x) = a(x – 3)(x – r2)

Now, to determine the value of r2, we need additional information. One way to proceed is to use the fact that the vertex form of a quadratic equation provides the x-coordinate of the vertex using the formula x = -b/2a, where a and b are the coefficients of the quadratic equation.

Since we have the x-coordinate of the vertex (3), we can express this equation as:

3 = -b/2a

Note: We are not given the specific value of a, so we will keep it as a variable.

Solving this equation for b, we get:

b = -6a

Now we substitute the values of r1 = 3 and b = -6a into the quadratic equation:

f(x) = a(x – 3)(x – r2)

If we expand the equation and collect like terms, it becomes:

f(x) = ax^2 – (3a + r2)a + 3r2a

For the quadratic to have distinct roots, the discriminant (b^2 – 4ac) must be greater than 0. In this case, expanding the discriminant gives:

(3a + r2)^2 – 4a(3r2a) > 0

9a^2 + 6ar2 + r2^2 – 12a^2r2 > 0

3a^2 + 6ar2 + r2^2 > 0

Now we need to consider different values for a and r2 that satisfy this inequality. Here’s an example:

Consider a = 1 and r2 = 2. Substituting these values into the quadratic equation, we have:

f(x) = (x – 3)(x – 2)

This equation represents a quadratic function in factored form with a vertex at x = 3 and two distinct roots.

Please note that this is just one example, and there are multiple combinations of a and r2 that can satisfy the criteria.

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