SOLUTION: 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 t

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: 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 t      Log On


   

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?
Answer by jim_thompson5910(35256) 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=%28-b%29%2F%282a%29.


x=%28-b%29%2F%282a%29 Start with the given formula.


From y=x%5E2%2Bx-2, we can see that a=1, b=1, and c=-2.


x=%28-%281%29%29%2F%282%281%29%29 Plug in a=1 and b=1.


x=%28-1%29%2F%282%29 Multiply 2 and 1 to get 2.


So the x-coordinate of the vertex is x=-1%2F2. Note: this means that the axis of symmetry is also x=-1%2F2.


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


y=x%5E2%2Bx-2 Start with the given equation.


y=%28-1%2F2%29%5E2%2B-1%2F2-2 Plug in x=-1%2F2.


y=1%2F4-1%2F2-2 Square -1%2F2 to get 1%2F4.


y=1%2F4-2%2F4-2 Multiply 1%2F2 by 2%2F2 to get 2%2F4.


y=1%2F4-2%2F4-8%2F4 Multiply 2 by 4%2F4 to get 8%2F4.


y=-9%2F4 Combine the fractions.


So the y-coordinate of the vertex is y=-9%2F4.


So the vertex is or in decimal form.