What is the vertex of #y=-x^2-3x+9 #?

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Answer 1 · 2018-06-26T03:58:02

Vertex: #(-1.5, 11.25)#

#y = -x^2 - 3x + 9#

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# and #b = -3#, so let's plug them into the formula:
#x = (-(-3))/(2(-1)) = 3/-2 = -1.5#

To find the #y#-coordinate of the vertex, just plug in the #x#-coordinate back into the original equation:
#y = -(-1.5)^2 - 3(-1.5) + 9#

#y = -2.25 + 4.5 + 9#

#y = 11.25#

Therefore, the vertex is at #(-1.5, 11.25)#.

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

As you can see, the vertex is indeed at #(-1.5, 11.25)#.

Hope this helps!