Questions on Algebra: Quadratic Equation answered by real tutors!

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Question 151429: I am trying to find the vertex for the function: f(x)=x^2 + x - 2. I believe the axis of symmetry is -1/2 since x= -b/2(a)(which should be the x-coordinate) but when I plug it in to find y I am coming up with - 3/2 which doesnt seem correct. Where am i going wrong, and what would be the vertex? : I am trying to find the vertex for the function: f(x)=x^2 + x - 2. I believe the axis of symmetry is -1/2 since x= -b/2(a)(which should be the x-coordinate) but when I plug it in to find y I am coming up with - 3/2 which doesnt seem correct. Where am i going wrong, and what would be the vertex?
Answer by jim_thompson5910(9217) About Me  (Show Source):
You can put this solution on YOUR website!

In order to find the vertex, we first need to find the x-coordinate of the vertex.


To find the x-coordinate of the vertex, use this formula: x=(-b)/(2a).


x=(-b)/(2a) Start with the given formula.


From y=x^2+x-2, we can see that a=1, b=1, and c=-2.


x=(-(1))/(2(1)) Plug in a=1 and b=1.


x=(-1)/(2) Multiply 2 and 1 to get 2.


So the x-coordinate of the vertex is x=-1/2. Note: this means that the axis of symmetry is also x=-1/2.


Now that we know the x-coordinate of the vertex, we can use it to find the y-coordinate of the vertex.


y=x^2+x-2 Start with the given equation.


y=(-1/2)^2+-1/2-2 Plug in x=-1/2.


y=1/4-1/2-2 Square -1/2 to get 1/4.


y=1/4-2/4-2 Multiply 1/2 by 2/2 to get 2/4.


y=1/4-2/4-8/4 Multiply 2 by 4/4 to get 8/4.


y=-9/4 Combine the fractions.


So the y-coordinate of the vertex is y=-9/4.


So the vertex is or in decimal form.