How do you graph $f (x) = \sqrt{x - 3} + 4$ $f(x)=sqrt(x-3)+4$ ?

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How do you graph $f (x) = \sqrt{x - 3} + 4$ $f(x)=sqrt(x-3)+4$ ?

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Answer 1 · 2015-02-03T16:40:10

First of all I have to be careful with the square root. I do not want a negative argument, so I set:
$(x - 3) \geq 0$ $(x-3)>=0$ and so:
$x \geq 3$ $x>=3$
This means that I can only give values of $x$ $x$ starting from $3$ $3$ and going up (I cannot give $2$ $2$ , $1$ $1$ , $0$ $0$ etc..)

Then I'll choose "friendly" values of $x$ $x$ in the sense that using them I can easily evaluate the square root:

For example:
$x = 3$ $x=3$ $\to$ $->$ $y = 0 + 4 = 4$ $y=0+4=4$
$x = 4$ $x=4$ $\to$ $->$ $y = 1 + 4 = 5$ $y=1+4=5$
$x = 7$ $x=7$ $\to$ $->$ $y = 2 + 4 = 6$ $y=2+4=6$
$x = 12$ $x=12$ $\to$ $->$ $y = 3 + 4 = 7$ $y=3+4=7$
...etc

I can now plot my graph:
enter image source here

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How do you graph $f (x) = \sqrt{x - 3} + 4$ $f(x)=sqrt(x-3)+4$ ?

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How do you graph f(x)=√x−3+4f(x)=sqrt(x-3)+4?

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How do you graph $f (x) = \sqrt{x - 3} + 4$ $f(x)=sqrt(x-3)+4$ ?