using vertex in graphing a quadratic function

Question

f(x) x^2-x-2  need help

Answer 1

The vertex of a quadratic in the form  ax^2 + bx + c  is located where x = -b/(2a)  

 

Here a = 1 and b = -1 so the vertex is located at x = 1/2 

 

f(1/2) = (1/2)^2 - (1/2) - 2 = 1/4 - 2/4 - 8/4 = -9/4  

 

So the vertex is located at (1/2,-9/4)  

 

The vertex is the turning point of the quadratic.  Since a = 1 is positive the quadratic opens upward.  The quadratic is symmetric about the vertical line x = 1/2 thru the vertex.  Plot some points near the vertex to fill out the graph  

 

f(0) = 0^2 - 0 - 2 = -2  so (0,-2) is on the graph

f(1) = 1^2 - 1 - 2 = -2 so (1,-2) is also on the graph.  Because the x-values 0 and 1 are symmetric about 1/2, these 2 y-values are the same 

 

Plot a few more points like f(2) and f(-1), and f(3) and f(-2) to complete the graph