How do you graph and identify the vertex and axis of symmetry for #5/2x(x-3)#?

Question

Eva Abramson · Accepted Answer

If #y=5/2x(x-3)# then you can put it into the form of #y=ax^2+bx+c# to create a table of points, then plot them on the graph.  But since you've already factorised it, you can find the solutions (x-intercepts)
#0=5/2x(x-3)#
#0=5/2x# or #0=x-3#
#x=0# or #x=3# The x-intercepts of your graph along with your vertex can allow you to plot a substantially accurate graph.  To find the vertex, complete the square:
#y=5/2x(x-3)#
#y=5/2x^2-15/2x#
#y=5/2(x^2-3x)#
#y=5/2(x^2-3x+(3/2)^2-(3/2)^2)#
#y=5/2((x-3/2)^2-9/4)#
#y=5/2(x-3/2)^2-45/8# Your vertex then becomes #(3/2, -45/8)#
which is the same as #(1.5, -5.625)#
And your axis of symmetry #x=1.5#

Avery Alexander · Answer

undefined To graph the quadratic function $ \frac{5}{2}x(x-3) $ and identify the vertex and axis of symmetry:

1. The vertex of a quadratic function in the form $ ax^2 + bx + c $ is given by the point $ \left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right) $.
2. The axis of symmetry is the vertical line passing through the vertex, given by the equation $ x = -\frac{b}{2a} $.

For the function $ \frac{5}{2}x(x-3) $:
- $ a = \frac{5}{2} $
- $ b = 0 $
- $ c = 0 $

So the vertex is $ \left( -\frac{0}{2\left(\frac{5}{2}\right)}, f\left(-\frac{0}{2\left(\frac{5}{2}\right)}\right) \right) = (0, 0) $.

The axis of symmetry is $ x = -\frac{0}{2\left(\frac{5}{2}\right)} = 0 $.

The graph is a parabola opening upwards, with the vertex at the origin (0,0), and the axis of symmetry is the y-axis.