How do you graph $f (x) = \sqrt{2 x + 3}$ $f(x)=sqrt(2x+3)$ ?

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How do you graph $f (x) = \sqrt{2 x + 3}$ $f(x)=sqrt(2x+3)$ ?

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Answer 1 · 2017-02-17T17:42:00

See explanation

Explanation:

Build a table of values relating $y$ $y$ to $x$ $x$ . You must take into account that what we really have is $y = \pm \sqrt{2 x + 3}$ $y=+-sqrt(2x+3)$ . Plot the points and draw the best line you can through those points.

Note that the x-intercept is where $y = 0 = \sqrt{2 x + 3}$ $y=0=sqrt(2x+3)$

Also note that $\sqrt{0} = 0 \Rightarrow 2 x + 3 = 0$ $sqrt(0)=0" "=>" "2x+3=0$

$2 x = - 3$ $2x=-3$

$x_{intercept} = - \frac{3}{2}$ $x_("intercept")=-3/2$
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Lets suppose you do not have a calculator. Then you need to select values for $2 x + 3$ $2x+3$ that are known squared values and adjust $x$ $x$ to give them.

This is the selection:
${(\pm 1)}^{2} = 1$ $(+-1)^2=1$
${(\pm 2)}^{2} = 4$ $(+-2)^2=4$
${(\pm 3)}^{2} = 9$ $(+-3)^2=9$
${(\pm 4)}^{2} = 16$ $(+-4)^2=16$
${(\pm 5)}^{2} = 25 \leftarrow$ $(+-5)^2=25 larr$ stop at this one as they are getting a bit big

$Set 2 x + 3 = 1 \Rightarrow 2 x = - 2 \Rightarrow x = - 1$ $color(blue)("Set "2x+3=1" " =>" " 2x=-2" " =>" "x=-1)$

$\pm y = \pm 1 = \pm \sqrt{2 (- 1) + 3}$ $" "+-y=+-1=+-sqrt(2(-1)+3)$
$\Rightarrow (x, y) = (- 1, + 1) & (- 1, - 1)$ $" "color(brown)(=> (x,y) = (-1,+1)" " &" " (-1,-1))$
................................................................................................
$Set 2 x + 3 = 4 \Rightarrow 2 x = + 1 \Rightarrow x = + \frac{1}{2}$ $color(blue)("Set "2x+3=4" " =>" " 2x=+1" " =>" "x=+1/2)$

$\pm y = \pm 2 = \pm \sqrt{2 (\frac{1}{2}) + 3}$ $" "+-y=+-2=+-sqrt(2(1/2)+3)$
$\Rightarrow (x, y) = (\frac{1}{2}, + 2) & (\frac{1}{2}, - 2)$ $" "color(brown)(=> (x,y) = (1/2,+2)" " &" " (1/2,-2))$

................................................................................................
$Set 2 x + 3 = 9 \Rightarrow 2 x = + 6 \Rightarrow x = + \frac{6}{2}$ $color(blue)("Set "2x+3=9" " =>" " 2x=+6" " =>" "x=+6/2$

$\pm y = \pm 3 = \pm \sqrt{2 (\frac{6}{2}) + 3}$ $" "+-y=+-3=+-sqrt(2(6/2)+3)$
$\Rightarrow (x, y) = (\frac{6}{2}, + 3) & (\frac{6}{2}, - 3)$ $" "color(brown)(=> (x,y) = (6/2,+3)" " &" " (6/2,-3))$

................................................................................................
$Set 2 x + 3 = 16 \Rightarrow 2 x = + 13 \Rightarrow x = + \frac{13}{2}$ $color(blue)("Set "2x+3=16" " =>" " 2x=+13" " =>" "x=+13/2$

$\pm y = \pm 4 = \pm \sqrt{2 (\frac{13}{2}) + 3}$ $" "+-y=+-4=+-sqrt(2(13/2)+3)$
$\Rightarrow (x, y) = (\frac{13}{2}, + 4) & (\frac{13}{2}, - 4)$ $" "color(brown)(=> (x,y) = (13/2,+4)" " &" " (13/2,-4))$

................................................................................................
$Set 2 x + 3 = 25 \Rightarrow 2 x = + 22 \Rightarrow x = + \frac{22}{2}$ $color(blue)("Set "2x+3=25" " =>" " 2x=+22" " =>" "x=+22/2$

$\pm y = \pm 5 = \pm \sqrt{2 (\frac{22}{2}) + 3}$ $" "+-y=+-5=+-sqrt(2(22/2)+3)$
$\Rightarrow (x, y) = (\frac{22}{2}, + 5) & (\frac{22}{2}, - 5)$ $" "color(brown)(=> (x,y) = (22/2,+5)" " &" " (22/2,-5))$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Your graph should look like this:

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How do you graph $f (x) = \sqrt{2 x + 3}$ $f(x)=sqrt(2x+3)$ ?

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