Graph the quadratic function f(x) = 3x² - 12x + 5, showing the vertex, axis of symmetry, and y-intercept.

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Let's graph the quadratic function f of x equals 3x squared minus 12x plus 5. First, we identify the key features. For a quadratic function in the form ax squared plus bx plus c, we have a equals 3, b equals negative 12, and c equals 5. Since a is positive, the parabola opens upward. To find the vertex, we use the formula negative b over 2a for the x-coordinate. This gives us x equals 2. Substituting this into the function, we get y equals negative 7. So the vertex is at the point (2, negative 7). The axis of symmetry passes through the vertex, so its equation is x equals 2. The y-intercept occurs when x equals 0, giving us the point (0, 5). Now, let's convert our quadratic function to vertex form. The standard form is f of x equals 3x squared minus 12x plus 5. To find the vertex form, we complete the square. First, we factor out the coefficient of x squared, which is 3. This gives us 3 times x squared minus 4x, plus 5. To complete the square inside the parentheses, we add and subtract the square of half the coefficient of x, which is 4. So we add and subtract 4. Rearranging, we get 3 times the quantity x squared minus 4x plus 4, minus 12 plus 5. The expression inside the parentheses is a perfect square trinomial, which equals x minus 2 squared. Simplifying further, we get f of x equals 3 times x minus 2 squared minus 7. This is the vertex form, where h equals 2 and k equals negative 7, confirming our vertex at the point (2, negative 7). Now, let's find the key points and intercepts of our quadratic function. The y-intercept occurs when x equals 0. Substituting x equals 0 into our function, we get f of 0 equals 3 times 0 squared minus 12 times 0 plus 5, which simplifies to 5. So the y-intercept is at the point (0, 5). To find the x-intercepts, we set f of x equals 0 and solve the equation 3x squared minus 12x plus 5 equals 0. Using the quadratic formula, x equals 12 plus or minus the square root of 144 minus 60, all divided by 6. This simplifies to 12 plus or minus the square root of 84, all divided by 6. Calculating these values, we get x approximately equals 0.47 or x approximately equals 3.53. So our x-intercepts are at the points (0.47, 0) and (3.53, 0). Notice how these points are symmetrically positioned around the axis of symmetry at x equals 2. Let's explore how our quadratic function f of x equals 3x squared minus 12x plus 5 can be understood as a series of transformations from the basic parabola y equals x squared. In vertex form, our function is f of x equals 3 times the quantity x minus 2 squared, minus 7. Starting with the basic parabola y equals x squared, we first apply a vertical stretch by a factor of 3, giving us y equals 3x squared. Next, we apply a horizontal shift 2 units to the right, resulting in y equals 3 times the quantity x minus 2 squared. Finally, we apply a vertical shift 7 units downward to get our original function: y equals 3 times the quantity x minus 2 squared, minus 7. These transformations explain why our vertex is at the point (2, negative 7): the x-coordinate 2 comes from the horizontal shift, and the y-coordinate negative 7 comes from the vertical shift. Understanding these transformations helps us visualize how the graph of our quadratic function relates to the basic parabola. Let's summarize what we've learned about the quadratic function f of x equals 3x squared minus 12x plus 5. The vertex of this parabola is at the point (2, negative 7), which we found using the formula negative b over 2a for the x-coordinate. In vertex form, the function can be written as f of x equals 3 times the quantity x minus 2 squared, minus 7, which shows how it's transformed from the basic parabola y equals x squared. The axis of symmetry is the vertical line x equals 2, passing through the vertex. The y-intercept is at the point (0, 5), and the x-intercepts are at approximately (0.47, 0) and (3.53, 0). Since the coefficient a equals 3 is positive, the parabola opens upward. Understanding these key features helps us fully characterize the graph of this quadratic function.