g(y) = 2(y^2 + 2y + 1) - 5(y + 1) + 1 = 2y^2 + 4y + 2 - 5y - 5 + 1 = 2y^2 - y - 2.
["# Simplifying the Quadratic Function ( g(y) = 2(y^2 + 2y + 1) - 5(y + 1) + 1 ): A Step-by-Step Analysis", "Quadratic functions are fundamental in algebra, modeling various real-world phenomena ranging from physics to economics. Today, we delve into simplifying the quadratic expression ( g(y) = 2(y^2 + 2y + 1) - 5(y + 1) + 1 ), transforming it into a more manageable form: ( g(y) = 2y^2 - y - 2 ). This simplification not only clarifies the structure of the function but also aids in graphing, finding roots, and analyzing its behavior.", "## The Original Form: Expanding ( g(y) )", "Let’s start by fully expanding the original expression step by step.", "[\n\begin{align}\ng(y) &= 2(y^2 + 2y + 1) - 5(y + 1) + 1 \\n&= 2y^2 + 4y + 2 - 5y - 5 + 1\n\end{align}\n]", "Combining like terms:", "- ( y^2 ) term: ( 2y^2 )\n- ( y ) terms: ( 4y - 5y = -y )\n- Constant terms: ( 2 - 5 + 1 = -2 )", "Thus,\n[\ng(y) = 2y^2 - y - 2\n]", "## Why Simplify Quadratic Functions?", "Simplifying quadratics improves clarity and makes it easier to:", "- Identify symmetry and vertex\n- Solve for roots using the quadratic formula or factoring\n- Determine the direction and width of the parabola\n- Graph the function accurately", "## Standard Form of a Quadratic Function", "The simplified expression ( g(y) = 2y^2 - y - 2 ) is now in standard quadratic form:", "[\ng(y) = ay^2 + by + c\n]", "where:\n- ( a = 2 ) (positive, so the parabola opens upwards)\n- ( b = -1 )\n- ( c = -2 )", "## Finding the Vertex: Axis of Symmetry", "The axis of symmetry, which passes through the vertex of the parabola, is given by:", "[\ny = -\frac{b}{2a} = -\frac{-1}{2 \cdot 2} = \frac{1}{4}\n]", "This tells us that the vertex lies at ( y = \frac{1}{4} ).", "## Calculating the Vertex (y-coordinate)", "Substitute ( y = \frac{1}{4} ) into ( g(y) ) to find the corresponding output:", "[\ng\left(\frac{1}{4}\right) = 2\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right) - 2\n]\n[\n= 2 \cdot \frac{1}{16} - \frac{1}{4} - 2\n]\n[\n= \frac{1}{8} - \frac{1}{4} - 2 = \left(\frac{1}{8} - \frac{2}{8}\right) - 2 = -\frac{1}{8} - 2 = -\frac{17}{8}\n]", "So, the vertex is at ( \left( \frac{1}{4}, -\frac{17}{8} \right) ).", "## Finding the Roots (x-intercepts)", "To find where the function crosses the y-axis (( g(y) = 0 )):", "[\n2y^2 - y - 2 = 0\n]", "Apply the quadratic formula:", "[\ny = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 16}}{4} = \frac{1 \pm \sqrt{17}}{4}\n]", "Thus, the roots are:", "[\ny = \frac{1 + \sqrt{17}}{4} \quad \ ext{and} \quad y = \frac{1 - \sqrt{17}}{4}\n]", "These values define the points where the parabola intersects the x-axis.", "## Summary of Key Features", "| Feature | Value/Description |\n|------------------------|-----------------------------------------|\n| Simplified Form | ( g(y) = 2y^2 - y - 2 ) |\n| Leading Coefficient ( a ) | 2 (parabola opens upward) |\n| Vertex | ( \left( \frac{1}{4}, -\frac{17}{8} \right) ) |\n| Axis of Symmetry | ( y = \frac{1}{4} ) |\n| Roots (Zeros) | ( y = \frac{1 \pm \sqrt{17}}{4} ) |", "## Conclusion", "The simplification of ( g(y) = 2(y^2 + 2y + 1) - 5(y + 1) + 1 ) to ( g(y) = 2y^2 - y - 2 ) reveals an elegant quadratic profile. Understanding this simplified form enables deeper analysis of its graph, roots, vertex, and symmetry — essential for both academic study and practical application.", "Whether you're solving equations, modeling real-life data, or preparing for a math exam, mastering quadratic simplification and analysis is a powerful skill. With this step-by-step guide, you’re now equipped to confidently interpret and manipulate quadratic functions like a pro."]
