• Oct 16th 2008, 08:34 AM
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--
• Oct 16th 2008, 08:52 AM
galactus
$\displaystyle y=x^{2}-1$

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

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

Did you notice this is a difference of two squares.

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

If you want the y-intercept, set x = 0 and solve. If you want the x-interceot, set y=0 and solve.
• Oct 16th 2008, 08:57 AM
masters
Quote:

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).

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

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

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

Too fast for me Galactus!
• Oct 16th 2008, 09:09 AM
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...(Bigsmile)
• Oct 16th 2008, 09:16 AM
masters
Quote:

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...(Bigsmile)

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

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

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.
• Oct 16th 2008, 09:32 AM
WhoaBlackBetty
Thank you. I think I understand it now! (Ninja)
• Oct 16th 2008, 11:39 AM
WhoaBlackBetty
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 ....
• Oct 16th 2008, 07:01 PM
WhoaBlackBetty
Anyone...