f(y) &= 3(y^2 - 4y + 4) + 12y - 24 + 7 \\

["Understanding the Quadratic Function: y = f(x) = 3(y² – 4y + 4) + 12y – 24 + 7", "When faced with a mathematical function like ( f(y) = 3(y^2 - 4y + 4) + 12y - 24 + 7 ), it’s common to feel overwhelmed by its structure. However, simplifying and understanding this expression step-by-step unlocks a clearer view of the quadratic relationship it represents. In this SEO-optimized article, we’ll break down the function ( f(y) ), simplify it into standard quadratic form, analyze its key features, and explain how to visualize and use it effectively.", "---", "### Step 1: Simplify the Quadratic Expression", "Start with the original function:", "[\nf(y) = 3(y^2 - 4y + 4) + 12y - 24 + 7\n]", "First, expand the expression inside the parentheses:", "[\n3(y^2 - 4y + 4) = 3y^2 - 12y + 12\n]", "Now substitute back:", "[\nf(y) = (3y^2 - 12y + 12) + 12y - 24 + 7\n]", "Combine like terms:", "- (3y^2) remains as is.\n- For (y) terms: (-12y + 12y = 0) → zero!\n- For constant terms: (12 - 24 + 7 = -5)", "So, the simplified form is:", "[\nf(y) = 3y^2 - 5\n]", "---", "### Step 2: Identify the Standard Form of a Quadratic Function", "The simplified function ( f(y) = 3y^2 - 5 ) is now in standard quadratic form:", "[\nf(y) = ay^2 + by + c\n]", "Here:\n- (a = 3) (a positive coefficient means the parabola opens upwards)\n- (b = 0) (so there is no (y) linear term)\n- (c = -5) (the y-intercept)", "---", "### Step 3: Key Characteristics of the Function", "#### 1. Vertex\nSince (b = 0), the vertex lies directly at ( y = 0 ). Substituting ( y = 0 ) into ( f(y) ):", "[\nf(0) = 3(0)^2 - 5 = -5\n]", "Vertex: ( (0, -5) )", "#### 2. Axis of Symmetry\nWith vertex at ( y = 0 ), the axis of symmetry is the vertical line:", "[\ny = 0\n]", "#### 3. Direction and Width\nBecause ( a = 3 > 0 ), the parabola opens upward and has moderate width due to ( a ) being relatively small in magnitude.", "#### 4. Intercepts\n- Y-intercept: When ( y = 0 ), ( f(y) = -5 ) → ( (0, -5) )\n- X-intercepts: Solve ( 3y^2 - 5 = 0 )\n [\n 3y^2 = 5 \Rightarrow y^2 = \frac{5}{3} \Rightarrow y = \pm \sqrt{\frac{5}{3}}\n ]\n So x-intercepts are ( \left( \pm \sqrt{\frac{5}{3}}, 0 \right) )", "---", "### Step 4: Graphing the Quadratic Function", "Plotting ( f(y) = 3y^2 - 5 ):", "- Start at the vertex at ( (0, -5) )\n- The parabola curves upward symmetrically as ( y ) moves left or right\n- Passes through points like ( (-1, 3 \cdot 1 - 5 = -2) ), ( (1, -2) ), and ( (\sqrt{5/3}, 0) )", "This function is useful in modeling scenarios where output grows rapidly with deviations from the origin—such as idealized cost models or physics trajectories assuming no linear shift.", "---", "### Step 5: Why Understanding f(y) Matters", "Simplifying complex algebraic expressions like the one given allows you to:", "- Easily identify the shape and key features of the graph\n- Solve for roots (x-values where output is zero)\n- Determine maximum/minimum values\n- Apply the function in applications like optimization or curve fitting", "Even though this function lacks ( y ) as both input and variable (a common point of confusion), recognizing it correctly is crucial in algebra, calculus, and applied mathematics.", "---", "### Summary", "The function ( f(y) = 3(y^2 - 4y + 4) + 12y - 24 + 7 ) simplifies elegantly to:", "[\nf(y) = 3y^2 - 5\n]", "This parabola opens upward with vertex at ( (0, -5) ), no linear term, and is centered on the y-axis. Understanding this simplified form supports better graphing, root-finding, and real-world application.", "---", "Keywords:\nf(y) function simplification, quadratic function, standard form, vertex form, parabola graph, find roots, simplify algebra, y = 3(y² – 4y + 4) + 12y – 24 + 7, 3y² - 5, parabola equation, algebra tutorial", "Meta Description:\nLearn how to simplify and analyze the quadratic function ( f(y) = 3(y^2 - 4y + 4) + 12y - 24 + 7 ) step-by-step, identifying its standard form, vertex, axis of symmetry, and graph behavior. Perfect for algebra students and math enthusiasts.", "---", "Optimized for search engines and clarity, mastering such functions builds strong foundational math skills essential in higher-level courses and real-world problem-solving."]