SOLUTION: Find the vertex and intercepts for the quadratic function y=x^2+x-6. Sketch the graph, and state the domain and range. Any help is greatly appreciated.

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Find the vertex and intercepts for the quadratic function y=x^2+x-6. Sketch the graph, and state the domain and range. Any help is greatly appreciated.      Log On


   

Question 7124: Find the vertex and intercepts for the quadratic function y=x^2+x-6. Sketch the graph, and state the domain and range. Any help is greatly appreciated.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
To find the x-intercepts (the zeros), set the quadratic = 0 and solve for x:
x%5E2+%2B+x+-+6+=+0 Factor.
%28x+%2B+3%29%28x+-+2%29+=+0 Apply the zero product principle.
x+%2B+3+=+0, x+=+-3
x+-+2+=+0, x+=++2
The x-intercepts are: (-3, 0) and (2, 0)
The x-coordinate of the vertex is at: -b/2a = -1/2. (ax^2 + bx + c = 0)Substitute this into the quadratic equation and solve for the y-coordinate.
y+=+%28-1%2F2%29%5E2+%2B+%28-1%2F2%29+-+6
y+=+1%2F4+-+1%2F2+-+6
y+=+-26%2F4 = - 6 1/4
The vertex is at: (-1/2, -6 1/4)
The graph of this quadratic equationis:
graph%28300%2C200%2C-5%2C5%2C-7%2C10%2Cx%5E2%2Bx-6%29
The domain (the set all possible values of the independent variable, that's x) is the set of all real numbers.
The range (the set of all possible values of the dependent variable, that's y) is also the set of all real numbers.