SOLUTION: Find the vertex for the graph of the quadratic function. (f)x= x^2 -13

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Question 171912: Find the vertex for the graph of the quadratic function.
(f)x= x^2 -13
Found 2 solutions by jim_thompson5910, checkley77:
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%5E2-13, we can see that a=1, b=0, and c=-13.


x=%28-%280%29%29%2F%282%281%29%29 Plug in a=1 and b=0.


x=%28-0%29%2F%282%29 Multiply 2 and 1 to get 2.


x=0 Divide.


So the x-coordinate of the vertex is x=0. Note: this means that the axis of symmetry is also x=0.


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


y=x%5E2-13 Start with the given equation.


y=%280%29%5E2-13 Plug in x=0.


y=0-13 Square 0 to get 0.


y=-13 Combine like terms.


So the y-coordinate of the vertex is y=-13.


So the vertex is .

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
(f)x=x^2-13
+graph%28+300%2C+200%2C+-6%2C+5%2C+-15%2C+10%2C+x%5E2+-13%29+ (graph 300x200 pixels, x from -6 to 5, y from -15 to 10, x^2 -13).
Vertex=(0,-13)