1. Let's start by understanding the problem: You want to know the vertex form, standard form, and factored form of a quadratic equation, which is a key topic in Grade 10 algebra. 2. The **standard form** of a quadratic equation is: $$y = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. 3. The **vertex form** is: $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex of the parabola. 4. The **factored form** is: $$y = a(x - r_1)(x - r_2)$$ where $r_1$ and $r_2$ are the roots (solutions) of the quadratic. 5. To convert from standard form to vertex form, use the method of completing the square: Example: Convert $$y = x^2 + 6x + 5$$ to vertex form. 6. Start with the standard form: $$y = x^2 + 6x + 5$$ 7. Group the $x$ terms: $$y = (x^2 + 6x) + 5$$ 8. Complete the square inside the parentheses: Take half of 6, which is 3, then square it: $3^2 = 9$. Add and subtract 9 inside the parentheses: $$y = (x^2 + 6x + 9 - 9) + 5$$ 9. Rewrite as: $$y = (x + 3)^2 - 9 + 5$$ 10. Simplify: $$y = (x + 3)^2 - 4$$ So, the vertex form is: $$y = (x + 3)^2 - 4$$ with vertex at $(-3, -4)$. 11. To find the factored form, solve for roots using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For $y = x^2 + 6x + 5$, $a=1$, $b=6$, $c=5$. Calculate the discriminant: $$\Delta = 6^2 - 4 \times 1 \times 5 = 36 - 20 = 16$$ Roots: $$x = \frac{-6 \pm \sqrt{16}}{2} = \frac{-6 \pm 4}{2}$$ So, $$x_1 = \frac{-6 + 4}{2} = -1$$ $$x_2 = \frac{-6 - 4}{2} = -5$$ 12. The factored form is: $$y = (x + 1)(x + 5)$$ Summary: - Standard form: $$y = x^2 + 6x + 5$$ - Vertex form: $$y = (x + 3)^2 - 4$$ - Factored form: $$y = (x + 1)(x + 5)$$ This process applies to any quadratic equation.