The function $ W(t) = -2t^2 + 12t + 50 $ is a quadratic in standard form $ at^2 + bt + c $, with $ a

["Understanding Quadratic Functions: Analyzing $ W(t) = -2t^2 + 12t + 50 $", "Quadratic functions are fundamental in algebra and appear in countless real-world applications, from projectile motion to business modeling. One such function is $ W(t) = -2t^2 + 12t + 50 $, which models a scenario with a quadratic relationship between time $ t $ and outcome $ W(t) $. In this article, we’ll explore the standard form of this quadratic, its key features, and how to analyze it using its algebraic structure.", "### What Makes $ W(t) = -2t^2 + 12t + 50 $ a Quadratic Function?", "A quadratic function is defined by the standard form:\n$$\nW(t) = at^2 + bt + c\n$$\nwhere $ a $, $ b $, and $ c $ are constants and $ a <br/>\neq 0 $. In our example:\n- $ a = -2 $\n- $ b = 12 $\n- $ c = 50 $", "Since the coefficient of $ t^2 $ is non-zero and constant, $ W(t) $ qualifies as a quadratic function. Its downward-opening parabola (due to the negative $ a $) indicates a maximum point, which has important implications for optimization and modeling.", "### Identifying Key Features of the Quadratic", "#### Vertex: The Peak of the Function\nThe vertex of a parabola described by $ at^2 + bt + c $ occurs at:\n$$\nt = -\frac{b}{2a}\n$$\nSubstituting $ a = -2 $ and $ b = 12 $:\n$$\nt = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3\n$$\nSo, the maximum value occurs at $ t = 3 $.\nTo find the maximum output $ W(3) $, substitute $ t = 3 $ into the original function:\n$$\nW(3) = -2(3)^2 + 12(3) + 50 = -18 + 36 + 50 = 68\n$$\nThus, the vertex is $ (3, 68) $, indicating the peak temperature, height, or value depending on the context.", "#### Axis of Symmetry\nThe vertical line $ t = 3 $ is the axis of symmetry — the graph is symmetric about this line. Any point on one side of $ t = 3 $ mirrors across this line to the corresponding point on the other side.", "#### Roots (Zeros) of the Function\nTo find when $ W(t) = 0 $, solve:\n$$\n-2t^2 + 12t + 50 = 0\n$$\nDivide through by $ -2 $ to simplify:\n$$\nt^2 - 6t - 25 = 0\n$$\nUse the quadratic formula:\n$$\nt = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-25)}}{2(1)} = \frac{6 \pm \sqrt{36 + 100}}{2} = \frac{6 \pm \sqrt{136}}{2}\n$$\nSince $ \sqrt{136} = 2\sqrt{34} $, the roots are:\n$$\nt = \frac{6 \pm 2\sqrt{34}}{2} = 3 \pm \sqrt{34}\n$$\nThese are real and distinct roots, meaning the function crosses the $ t $-axis at two points, reflecting the quadratic’s ability to model crossing values.", "### Practical Applications of $ W(t) $\nQuadratic functions like $ W(t) $ model phenomena where growth changes direction, such as:\n- Projectile motion (maximum height at peak time)\n- Revenue with price-sensitive demand\n- Area maximization in architectural design", "Understanding $ W(t) $’s shape and location helps predict outcomes, optimize decisions, and interpret real data.", "### Summary of Coefficients Significance\n- $ a = -2 $: Determines the parabola’s concavity (opens downward), scale, and steeper curvature.\n- $ b = 12 $: Influences the horizontal shift relative to the vertex.\n- $ c = 50 $: The $ y $-intercept, representing the initial value of $ W(t) $ when $ t = 0 $.", "---", "Conclusion\nThe function $ W(t) = -2t^2 + 12t + 50 $ exemplifies a downward-opening quadratic with clear real-world relevance. By analyzing $ a $, $ b $, $ c $, and applying vertex and symmetry, we uncover its peak behavior and full graphic properties. Whether used in science, economics, or engineering, mastering such quadratic forms strengthens problem-solving across disciplines.", "Keywords: quadratic function, $ W(t) = -2t^2 + 12t + 50 $, standard form $ at^2 + bt + c $, vertex, axis of symmetry, quadratic roots, real-world applications, algebra."]