How to graph a parabola $y = \frac{3}{2} x^{2} - 3 x - \frac{5}{2}$ $y=3/2x^2 - 3x-5/2$ ?

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Answer 1 · 2015-06-21T20:49:50

Complete the square to find the vertex, axis of symmetry, etc.

$\frac{3}{2} {(x - 1)}^{2} = \frac{3}{2} (x^{2} - 2 x + 1) = \frac{3}{2} x^{2} - 3 x + \frac{3}{2}$ $3/2(x-1)^2 = 3/2(x^2-2x+1) = 3/2x^2-3x+3/2$

So

$y = \frac{3}{2} x^{2} - 3 x - \frac{5}{2}$ $y = 3/2x^2-3x-5/2$

$= \frac{3}{2} {(x - 1)}^{2} - \frac{3}{2} - \frac{5}{2}$ $= 3/2(x-1)^2-3/2-5/2$

$= \frac{3}{2} {(x - 1)}^{2} - 4$ $= 3/2(x-1)^2-4$

This is in vertex form,

allowing us to tell that the vertex is at $(1, - 4)$ $(1, -4)$

and the axis of symmetry is $x = 1$ $x=1$ .

The intersection with the $y$ $y$ axis is at $(0, \frac{3}{2})$ $(0, 3/2)$

The intersections with the $x$ $x$ axis are where $y = 0$ $y=0$ , so

$0 = \frac{3}{2} {(x - 1)}^{2} - 4$ $0 = 3/2(x-1)^2 - 4$

Add $4$ $4$ to both sides to get:

$\frac{3}{2} {(x - 1)}^{2} = 4$ $3/2(x-1)^2 = 4$

Multiply both sides by $\frac{2}{3}$ $2/3$ to get:

${(x - 1)}^{2} = \frac{2}{3} \cdot 4$ $(x-1)^2 = 2/3*4$

Hence $x = 1 \pm 2 \sqrt{\frac{2}{3}} ≅ 1 \pm 1.633$ $x = 1+-2sqrt(2/3) ~= 1+-1.633$

So the intersections with the $y$ $y$ axis are approximately:

$(- 0.633, 0)$ $(-0.633,0)$ and $(2.633, 0)$ $(2.633, 0)$

graph{3/2x^2-3x-5/2 [-10, 10, -5, 5]}

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