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--

2. $y=x^{2}-1$

The vertex is easier found by using $x=\frac{-b}{2a}$

In this case, a=1 and b=0, c=-1.

Did you notice this is a difference of two squares.

$y=x^{2}-1=(x+1)(x-1)$

If you want the y-intercept, set x = 0 and solve. If you want the x-interceot, set y=0 and solve.

3. Originally Posted by WhoaBlackBetty
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--
To find the y-intercept, find f(0).

$x^2-1=0$
$f(0)=0^2-1$
$f(0)=-1$. This is your y-intercept.

To find the x-intercepts (zeros of the function), set f(x)=0 and solve for x.

$x^2-1=0$
$(x-1)(x+2)=0$
$x=1 \ \ or \ \ x=-1$ These are your x-intercepts.

Too fast for me Galactus!

4. 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...

5. Originally Posted by WhoaBlackBetty
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...
Set up a table using arbitrary x values. Then find f(x) to complete the ordered pair. Then plot them.

$f(x)=x^2-1$

x=0, f(0)=-1, Plot (0, -1)
x=1, f(1)= 0, Plot (1, 0)
x=-1, f(-1)=0, Plot (-1, 0)
.
.
.
etc.

6. Thank you. I think I understand it now!

7. 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 ....

8. Anyone...