How do you find the axis of symmetry and vertex point of the function: #y= -1.5x^2+6x#?

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Answer 1 · 2018-06-22T20:56:32

Vertex: #(2, 6)#

Axis of symmetry: #x = 2#

#y = -1.5x^2 + 6x#

To find the #x#-coordinate of the vertex of a standard quadratic equation (y = ax^2 + bx + c), we use the formula #(-b)/(2a)#.

We know that #a = -1.5# and #b = 6#, so let's plug them into the formula:
#x = (-6)/(2(-1.5)) = -6/-3 = 2#

To find the #y#-coordinate of the vertex, just plug in the #x#-coordinate back into the original equation:
#y = -1.5x^2 + 6x#

#y = -1.5(2)^2 + 6(2)#

#y = -1.5(4) + 12#

#y = -6 + 12#

#y = 6#

Therefore, the vertex is at #(2, 6)#.

The axis of symmetry is the line of the #x#-coordinate of the vertex, so it's #x = 2#.

Here's a graph of this equation (desmos.com):
enter image source here

As you can see, the vertex is indeed at #(2, 6)#.

Hope this helps!