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A quadratic function is often written in a specific way called "standard form." This is usually expressed as \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. This form is very useful for identifying important features of the quadratic function like its vertex and intercepts.
In the exercise, the function \(f(x) = -3x^2 + 6x - 2\) is already in standard form, so there's no need to rewrite it. Here, the coefficients are:\- \(a = -3\)
- \(b = 6\)
- \(c = -2\)
Being comfortable with recognizing this form is step one in understanding and graphing quadratic functions. It provides a straightforward way to work with and analyze the parabola formed by the quadratic equation.

The vertex of a quadratic function is a critical point that provides insight into the parabola's maximum or minimum value. For the standard form quadratics \(ax^2 + bx + c\), the vertex can be calculated using \(h = -\frac{b}{2a}\).
Once you have \(h\), plug it into the function to find \(k\), which gives the vertex as \((h, k)\).
In our specific example, \(h\) is calculated as: \[h = -\frac{6}{2(-3)} = 1\]
The corresponding \(k\) value is found by substituting 1 back into the function \[f(1) = -3(1)^2 + 6 \times 1 - 2 = 1\]
This results in a vertex of \((1, 1)\).
The vertex provides a peak or a valley for the function—here, it's a peak since the quadratic opens downward (because \(a = -3\) is negative).

Finding the intercepts of a quadratic function means pinpointing where the graph crosses the axes. We have two types of intercepts: **x-intercepts** and **y-intercepts.**

x-intercepts are found by setting the quadratic equation equal to zero and solving for \(x\). For \(-3x^2 + 6x - 2 = 0\), the quadratic formula offers a solution: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Substituting in the values gives: \[x = \frac{-6 \pm \sqrt{36 - 24}}{-6} = 1 \pm \frac{\sqrt{3}}{3}\]
The x-intercepts are at \(1 + \frac{\sqrt{3}}{3}\) and \(1 - \frac{\sqrt{3}}{3}\).

y-intercepts, on the other hand, can be found by setting \(x = 0\) and solving for \(y\):
\[f(0) = -3(0)^2 + 6(0) - 2 = -2\]
This provides a y-intercept at the point \((0, -2)\).
These intercepts serve as markers on the graph, indicating where the parabola touches the axes.

Graphing a quadratic function is both an art and science. With some practice and the right points, you can sketch a precise representation of the function. Here's how you can do it:
- **Plot Key Points:** Start by plotting the vertex, which is the turning point of the parabola. In this case, the vertex is \((1, 1)\).
- **Mark Intercepts:** Add the x-intercepts and y-intercept to your graph. Here, x-intercepts are at \(1 + \frac{\sqrt{3}}{3}\) and \(1 - \frac{\sqrt{3}}{3}\), with a y-intercept at \((0, -2)\).
- **Determine Direction:** Since the function opens downward (because \(a = -3\) is negative), it will look like an upside-down "U."
- **Draw the Parabola:** Connect these points smoothly to form a U-shaped curve, making sure the parabola passes through all intercepts and that it curves smoothly at the vertex.
Using these steps allows you to visualize how the factors of the quadratic function translate into the graph's shape and direction. Graphing is an excellent way to understand how small changes to \(a\), \(b\), or \(c\) alter the entire graph.