Graphing

**pashah** · August 4th 2006, 05:11 AM

Hello I would appreciate any help with the following problem.

**galactus** · August 4th 2006, 05:49 AM

The graph of $\sqrt{x+5}-2$

Can you see what the domain is?.

**Quick** · August 4th 2006, 05:54 AM

quick overview: hopefully you know what this $y=\sqrt{x}$ looks like,

now when you see an equation like this $y=\sqrt{x+5}$ you move the graph of $y=\sqrt{x}$ left 5 units, if you see an equation like this $y=\sqrt{x}-2$ you move the graph of $y=\sqrt{x}-2$ down 2 units.

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

the x-intercepts happen when f(x)=0 therefore we have: $0=(x+2)(x-2)^2$

now anything multiplies by zero is zero, so only one of the two groups has to be zero for the entire equation to go, so we figure out when either of them are zero:

$\begin{array}{cc}x+2=0\Rightarrow x=-2\\x-2=0\Rightarrow x=2\end{array}$

so the y-intercepts are when $x=\pm2$

**Soroban** · August 4th 2006, 06:32 AM

Hello, pashah!

Sketch the graph and state the domain and range.

. . $y \;= \;\sqrt{x+5} - 2$

Since $x + 5$ must be nonnegative: . $x \geq -5$ (domain)

The minimum value of $y$ occurs when $x = -5$
Hence, the range is: . $y \geq -2$

We have: . $y + 2 \:=\:\sqrt{x+5}$
Square the equation: . $(y+2)^2 \:=\:x + 5$
We have a parabola with vertex $(-5,-2)$ which opens to the right: $\subset$
And we want only the upper half.

Code:

                  |
                  |      *
                  *
      - - - - * - + - - - - -
           *      |
         *        |
        *         |
     (-5,-2)      |

Graph and identity all intercepts.

. . $y \;= \;(x+2)(x-2)^2$

If $x = 0,\;y\:=\:2(-2)^2 \:=\:8$ . . . y-intercept: $(0,8)$

If $y = 0,\;x\:=\:\pm2$ . . . x-intercepts: $(\pm2,0)$

This is a cubic graph which rises to the right.
Since $(x-2)^2$ has an even exponent,
. . the graph is tangent to the x-axis at $x = 2$

Code:

                  |
                  * *         *
           -2  *  |   *     *
      - - - * - - + - - -*- - -
          *       |      2
         *        |

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