# Given the function f(x)=-3x^2+6x-4. Draw the grap and determine the vertex, axis of symmetry, y-intercept and find an additional point on graph.

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You need to determine the vertex of the graph and then you may tell what the axis of symmetry is.

You need to use the coefficients of the quadratic to find the coordinates of vertex such that:

`x_V = -b/(2a) ; y_V = (4ac - b^2)/(4a)`

Substituting a=-3 ; b = 6 ; c = -4 yields

`x_V = -6/(-6) = 1`

`y_V = (48-36)/(-12) = 12/(-12) = -1`

Hence, the coordinates of the vertex are (1,-1) and the axis of symmetry passes through the vertex.

Since the leading coefficient is negative, the parabola is co ncave down, hence the graph does not intercept x axis.

The point where the graph intercepts y axis may be considered the additional point on the graph, hence you need to substitute 0 for x such that:

f(0) = -4

Hence, evaluating the additional point on the graph yields (0,-4).

Sketching the graph of the function yields:

**Notice the coordinates of vertex (1,-1) and the point (0,-4) where the graph intercepts y axis.**

The function f(x)=-3x^2+6x-4. The graph of the function is

The vertex of the graph is (1, -1). The axis of symmetry is x = 1. The y-intercept is (0, -4) and an additional point on the graph is (2, -4).