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graph y=-3/4x^2 plot the vertex and four additional points, two on each side of the vertex. |

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Graph `y=(-3)/4x^2` :

(1) This is a quadratic function so the graph is a parabola.

The leading coefficient is negative so it opens down.

The absolute value of the leading coefficient is less than 1, so the parabola opens "wider" than the standard.

(2) The axis of symmetry is the line `x=(-b)/(2a)` ; since b=0 the axis is x=0;

The vertex lies on the axis of symmetry so the vertex is at `(0,f(0))` or (0,0).

(3)The graph is symmetric across the line x=0, so f(a)=f(-a). Plugging in some values for x we get:

`f(1)=(-3)/4` so by symmetry `f(-1)=(-3)/4`

`f(2)=-3` so by symmetry `f(-2)=-3`

`f(4)=-12` so by symmetry `f(-4)=-12`

Thus we have the points (-4,-12),(-2,-3),(-1,-3/4),(0,0),(1,-3/4),(2,-3),(4,-12)

The graph:

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