The given function is:

$F(x) = x^2 - 8x + 12$

Step 1: Factorization

To factorize $F(x)$, we look for two numbers that multiply to $12$ (the constant term) and add up to $-8$ (the coefficient of $x$).

The numbers are $-6$ and $-2$. So, we can rewrite the quadratic as: $F(x) = (x - 6)(x - 2)$

Step 2: Roots

Setting $F(x) = 0$ gives: $(x - 6)(x - 2) = 0$

The solutions (roots) are: $x = 6 \quad \text{and} \quad x = 2$

Step 3: Vertex (Turning Point)

The vertex of a quadratic function $ax^2 + bx + c$ is given by: $x = -\frac{b}{2a}$ Here, $a = 1$, $b = -8$. Substituting: $x = -\frac{-8}{2(1)} = 4$

To find $y$-coordinate of the vertex: $F(4) = (4)^2 - 8(4) + 12 = 16 - 32 + 12 = -4$

The vertex is: $(4, -4)$

Step 4: Axis of Symmetry

The axis of symmetry is the vertical line passing through the vertex: $x = 4$

Step 5: Graph Features

• Opening: Since $a = 1 > 0$, the parabola opens upwards.
• Y-intercept: Substitute $x = 0$ into $F(x)$: $F(0) = 0^2 - 8(0) + 12 = 12$ Thus, the y-intercept is $(0, 12)$.

Summary

1. Factored form: $F(x) = (x - 6)(x - 2)$
2. Roots: $x = 6, x = 2$
3. Vertex: $(4, -4)$
4. Axis of symmetry: $x = 4$
5. Y-intercept: $(0, 12)$

Would you like me to graph this function or explain any step in more detail?

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Here are some related questions:

1. How can the vertex form of $F(x)$ be derived?
2. What is the significance of the axis of symmetry in quadratic graphs?
3. How do transformations affect the standard quadratic function $y = x^2$?
4. Can you explain how to find the domain and range of $F(x)$?
5. How would $F(x)$ change if a constant is added or subtracted?

Tip: Always check your factored forms by expanding them back to the original quadratic equation!