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blacksinger · Jan 5, 2009, 09:53 PM

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.

Unknown008 · Jan 6, 2009, 01:57 AM

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?
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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? :)