# Help with graphs

• Oct 20th 2007, 03:45 AM
Kblack
Help with graphs
Simplify the expressions:

Pls check attachment

2) a) Find the equation of a circle centered at the point (3,4) with radius 5.
b) Is the point (0,0) on the above circle ?

3) Find the domain of the functions
f(x)=sqrt16-x2
g(x) = 1/x2-4

4) a) Graph the function f(x)=x2-2
b) Find the equation of the function whose graph is the graph of f, shifted by 3 to the right.
c) Find the equation of the function whose graph is the graph of f, shifted up by 3.

5) a) Find the two roots (solutions) of the quadratic equation
(x+1)2 + 4 = 0 (hint: they are complex numbers)
b) Calculate the product of the two roots.
• Oct 20th 2007, 04:36 AM
topsquark
Quote:

Originally Posted by Kblack
2) a) Find the equation of a circle centered at the point (3,4) with radius 5.
b) Is the point (0,0) on the above circle ?

The formula for a circle is
$(x -h)^2 + (y - k)^2 = r^2$
where (h, k) are the coordinates for the center of the circle and r is the radius.

To find out if (0, 0) is on the circle, just plug in x = 0 and y = 0 into your circle equation and see if the equation reduces to a true statement (such as 25 = 25).

-Dan
• Oct 20th 2007, 04:39 AM
topsquark
Quote:

Originally Posted by Kblack
3) Find the domain of the functions
f(x)=sqrt16-x2
g(x) = 1/x2-4

Please use parenthesis!! For example, "f(x) = sqrt16 - x2" means $f(x) = \sqrt{16} - x^2 = 4 - x^2$.

So.
$f(x) = \sqrt{16 - x^2}$
Recall that you can't have a negative number under the radical. Thus we have the condition that $16 \geq x^2$. You take it from here.

$g(x) = \frac{1}{x^2 - 4}$
The denominator of an expression cannot be 0. For what x values is this expression 0?

-Dan
• Oct 20th 2007, 04:44 AM
topsquark
Quote:

Originally Posted by Kblack
4) a) Graph the function f(x)=x2-2
b) Find the equation of the function whose graph is the graph of f, shifted by 3 to the right.
c) Find the equation of the function whose graph is the graph of f, shifted up by 3.

I'll let you do the graphing part.

When you shift an equation y = f(x) h units to the left, you replace x with x - h. When you shift it h units to the right, you replace x with x + h. (Yes, this seems backward. I recommend you graph both of these to see for yourself.)

When you shift and equation y = f(x) k units upward, you replace y with y - k. When you shift it k units downward, you replace y with y - k.

So for example you start with $y = f(x) = x^2 - 2$ and you want to shift it 3 units to the right. Then you replace x with x + 3, so we have a "new" function $y = f(x + 3) = (x + 3)^2 - 2 = x^2 + 6x + 7$.

You do the other case.

-Dan
• Oct 20th 2007, 04:48 AM
topsquark
Quote:

Originally Posted by Kblack
5) a) Find the two roots (solutions) of the quadratic equation
(x+1)2 + 4 = 0 (hint: they are complex numbers)
b) Calculate the product of the two roots.

$(x + 1)^2 + 4 = 0$

We aren't told how to solve it, so I'll solve it this way:
$(x + 1)^2 = -4$

$x + 1 = \pm \sqrt{-4} = \pm i\sqrt{4} = \pm 2i$

$x = -1 \pm 2i$

So the solutions for x are $x = -1 + 2i$ and $x = -1 - 2i$.

To get their product, FOIL it out:
$(-1 + 2i)(-1 - 2i) = (-1)(-1) + (-1)(-2i) + (2i)(-1) + (2i)(-2i)$

I'll let you finish it from here. Recall that $i^2 = -1$.

-Dan
• Oct 20th 2007, 05:02 AM
topsquark
Quote:

$(x^6)^{1/2}$

$(\sqrt{x})^{1/3}$

$\left ( \frac{4}{9} \right ) ^{1/2} + \left ( \frac{1}{3} \right ) ^2$
Recall the properties of exponents:
$a^ma^n = a^{m + n}$

$a^{-n} = \frac{1}{a^n}$

$(a^m)^n = a^{mn}$

And also:
$\sqrt[n]{x} = x^{1/n}$
(So $\sqrt{x} = x^{1/2}$.)

You want to put these into the form: $a^m$. (Except for the last one, just give the fraction for the answer.) For the second one:
$(\sqrt{x})^{1/3} = \left ( x^{1/2} \right )^{1/3} = x^{1/6}$

See what you can do for the first one.

Quote:

$\frac{3}{5} + \frac{1}{6}$
You've got to be kidding me. You are using exponent laws and solving quadratic equations and you are saying you don't know how to add two fractions?? Get the least common multiple of 5 and 6, then put the fractions in terms of that as a denominator.

-Dan
• Oct 20th 2007, 10:03 PM
earboth
Quote:

Originally Posted by Kblack
...

4) a) Graph the function f(x)=x2-2
b) Find the equation of the function whose graph is the graph of f, shifted by 3 to the right.
c) Find the equation of the function whose graph is the graph of f, shifted up by 3.

...

Hello,

b) if you have to shift the graph of your function to the right by 3 units you are looking for a new function $f(\overline{x})=\overline{x}^2-2$
You know that $\overline{x} = x+3~\iff~x=\overline{x}-3$. Plug in the term of the RHS of this equation into your original equation and you'll get the equation of the translated graph:

$f(\overline{x})=\left(\overline{x}-3\right)^2-2$

c) if you have to shift the graph of your function 3 units up then you are looking for a new funktion $\overline{y} = f(x)$. You know that $\overline{y} = y+3~\iff~\overline{y} = f(x)+3$. Therefore the equation of the translated graph is:

$\overline{y} = x^2-2+3=x^2+1$

In general:
If you want to shift to the right by s units replace x by (x-s)
If you want to shift to the left by s units replace x by (x+s)
If you want to shift upward by s units add s to the term of the function
If you want to shift downward by s units add (-s) to the term of the function