A curve has equation y=x^2 - 6x + 14

Question

Write the equation in the form y=(x+a)^2 + b where a and b are integers. 

Draw a sketch of the curve showing clearly the coordinates of any turning points and where the curve crosses the y - axis.

Answer 1

y = x²  -6x + 14  this is standard form for a quadratic equation

                         its graph will be a parabola

 

Change this to Graphing form: y = a ( x - h )² + k

 

(h, k) will be your vertex point, 'a' has same value in either form

 

y = x² -6x + 14   complete the square using only the x terms

 

y = ( x -3 )²      this would make the constant term 9 and not 14

                      so add 5

 

y = ( x - 3 )² + 5  is the equation of a parabola with vertex (3,5)

 

Since a = 1, this parabola opens the same rate as y = x²

 

To find where the curve crosses the y-axis: use either form and 

letting x=0, solve for y

 

Answer 2

The form you have written down is incorrect.  It should be

 

y = (x - a)² + b   --->  vertex form of parabola.

 

 

y = (x² - 6x + 9) + 5

 

y = (x - 3)(x - 3) + 5

 

y = (x - 3)² + 5

 

 

Now to sketch the curve, plot any x-intercepts, the y-intercept, and the vertex.  Then connect the points in the path they follow.