The vertex is easier found by using
In this case, a=1 and b=0, c=-1.
Did you notice this is a difference of two squares.
If you want the y-intercept, set x = 0 and solve. If you want the x-interceot, set y=0 and solve.
Another function problem...
Suppose I have the function f(x)=x^2-1 how would I find the intercepts?
The vertex is easy to find by completing the square. y=a(x-h)^2+k where a= h=0 and k=-1 giving the location of the vertex at (0,-1) Could someone explain how I get them? I keep getting the wrong answer--
Bam bah Lam--
Ok, that makes more since. This is confusing for some reason to me. Harder than just solving the quadratic. So, if I was going to graph it how would I go about that? I know my first point the vertex would be at (0,-1) So, I would have a point there, but would I just use the x-intercepts combined with plugging in a Y value for the other two points? Or something different? Thanks...
One more, I am doing something wrong. I know what the values are supposed to be like (2,4) (3,1) (4,0) (5,1) (6,4)...etc. How do I plug the values in to get those numbers? It must be something simple I am missing. Here is the problem f(x)=(x-4)^2
This is taking me longer than usual to wrap my head around... So, I know this is a horizontal shift with the vertex at (4,0) The part I am not getting is plugging it into the equation. Take this for example--if
f(x)= (x-4)^2 with x= to 1
f(1)= (1-4)(1-4) Or, am I mathing these out wrong? Doing this way you wind up with 1 and 9 unless you take the square which is 1,3 ....