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

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Answer 1 · 2017-03-16T01:07:45

#(-1,4)#

There is a lovely and straightforward (which makes it all the lovelier) rule for working out vertices such as this one.

Think of the general parabola: #y=ax^2+bx+c#, where #a!=0#

The formula for finding the #x#-vertex is #(-b)/(2a)# and to find the #y#-vertex, you insert the value you found for #x# into the formula.

Using your question #y=-x^2-2x+3# we can establish the values of #a, b, #and #c#.

In this case:

#a=-1#
#b=-2#; and
#c=3#.

To find the #x#-vertex we need to replace the values for #a# and #b# in the formula given above (#color(red)((-b)/(2a))#):

#=(-(-2))/(2*(-1))=2/(-2)=-1#

So we now know that the #x#-vertex is at #-1#.

To find the #y#-vertex, go back to the original question and replace all the instances of #x# with #-1#:

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

#y=-(-1)^2-2*(-1)+3#

#y=-1+2+3#

#y=4#

We now know that the #x#-vertex is at #-1# and the #y#-vertex is at #4# and this can be written in coordinate format:

#(-1,4)#