f(x)=-x^2+2x-5

Question

determine the vertex

indemnify the interval of increase

rewrite the function in vertex form

Answer 1

For a quadratic in the form ax^2 + bx + c = 0 the vertex is located at (-b/(2a),f(-b/(2a))

 

Here a = -1, b = 2, so the vertex is located at x = -2/(2*(-1)) = 1

 

f(1) = -1 + 2 - 5 = -4  

 

So the vertex is at (1,-4) 

 

Since a is negative the quadratic opens downward, which means it is increasing on the left side of the vertex.  This is the interval 

 

(-∞,1) 

 

To rewrite the function in vertex form, first complete the square 

 

f(x) = -x^2 + 2x - 5 = -(x^2 - 2x + 5) = -(x^2 - 2x + 1 + 4) 

      = -(x-1)^2 - 4