Ex. f(x)=x2-7

How would you solve that? I know it has to do with this forumla:
y= a(x-h)2-3

My teacher had said to find the perfect square. Seven isn't a number has the two same numbers making it.


Please help.

From your given function, I would be able to give you the cordinates of the vertex directly. That's possible if you know the basic function 'y = x^2'. With y = x^2, adding a +1 (like this y = x^2 + 1) translates the entire graph up one unit. Similarly, putting a -1 will translate the original graph one unit down. Putting a +1 or -1 before putting the square sign, like y = (x+1)^2, y = (x-1)^2 will move the graph horizontally one unit to the left and one unit to the right, respectively.

In the function y = x^2, the vertex has the cordinates (0, 0).
In the function y = x^2 + 1, the vertex has cordinates (0, 1).
In the function y = x^2 - 1, the vertex has cordinates (0, -1).
In the function y = (x+1)^2, the vertex has cordinates (-1, 0).
In the function y = (x-1)^2, the vertex has cordinates (1, 0).

Now, can you find the cordinates of the vertex of your function?
----------

For the x-intercepts, equate your function to '0', and solve. If you have y=x^2 -3, you'll have to solve 0=x^2 -3 and obtain x=\pm{\sqrt3} as solution.

For the y-intercept, set 'x' to '0' and solve for 'y'.

Completing the square is for functions that are not in the format a(x-b)^2+c, but in for example ax^2 + bx + c, which cannot be factorised unless you use a formula.

Post back if you don't understand something, or need more help, OK? :)