# How do you find its vertex, axis of symmetry, y-intercept and x-intercept for #f(x) = -3x^2 + 3x - 2#?

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Find the axis of symmetry using the equation

Find the vertex by substituting the value for

There are no x-intercepts.

To get the y-intercept, substitute 0 for

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The general formula for a quadratic equation is

The graph of a quadratic equation is a parabola. A parabola has an axis of symmetry and a vertex. The axis of symmetry is a vertical line the divides the parabola into to equal halves. The line of symmetry is determined by the equation

**Axis of Symmetry**

The axis of symmetry is the line

**Vertex**

Determine the value for

The common denominator is

The vertex is

**X-Intercept**

The x-intercepts are where the parabola crosses the x-axis.There are no x-intercepts for this equation because the vertex is below the x-axis and the parabola is facing downward.

**Y-Intercept**

The y-intercept is where the parabola crosses the y-axis. To find the y-intercept, make

The y-intercept is

graph{y=-3x^2+3x-2 [-14, 14.47, -13.1, 1.14]}

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f(x) = - 3x^2 + 3x - 2

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x of vertex = x of axis of symmetry:

y of vertex:

y intercept: y = -2

x-intercepts -> y = 0

D = b^2 - 4ac = 9 - 24 = - 15 < 0. There are no real roots (no x-intercepts) because D < 0.

Since a < 0, the parabola opens downward. The parabola is completely below the x-axis.

Describe your changes (optional) 200