# Thread:

1. ## How do you graph this function?

$f(x)=x^2+6x$

2. Originally Posted by juliajank
$f(x)=x^2+6x$
There are alot of ways to do this the most simple is to plot points.

So pick some values of x and plug them in to get y values and plot the points...

here is one let $x=1 \implies f(1)=1^2+6(1)=7 \implies (1,7)$

3. Thanks a lot for your answer, but I can't use the table of values anymore

Do you have to factor out the common x first to make it
x(x+6) and then graph it as a normal line, six units left?
or are you supposed to graph it as a quadratic function? :S
When you factor the x out, does the make whatever is in the brackets the only important thing now?
I am so confused with what I have to do if there is more than one x!

4. Originally Posted by juliajank
Thanks a lot for your answer, but I can't use the table of values anymore

Do you have to factor out the common x first to make it
x(x+6) and then graph it as a normal line, six units left?
or are you supposed to graph it as a quadratic function? :S
When you factor the x out, does the make whatever is in the brackets the only important thing now?
I am so confused with what I have to do if there is more than one x!

There are many methods. What method are you supposed to use.

Here is one...

We could first find the x-intercepts by setting y=0

$0=x^2+6x=x(x+6) \implies x=0 or x=-6$

So the ordered pairs of the x intercepts are (0,0) and (-6,0)

Frome here we can find the vertex by using the fact that a parabola is symmetric about its vertex. That means the x coordinate of the vertex is at the mid point of the x intercepts. $x=\frac{0+(-6)}{2}=-3$

Now we can complete the order pair by plugging this back into the equation
$y=(-3)^2+6(-3)=9-18=-9$ so the vertex is at (-3,-9)

So if we plot these three points we can sketch the parabola.

Another methold we be to complete the square...

Do any of these sound familar...

5. Yes! Thank you. I would have to solve it by completing the square which is not very hard. It's just figuring out what to do that's trickiest for me.

6. Originally Posted by juliajank
It's just figuring out what to do that's trickiest for me.
For worked examples of the general methodology, try here.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9559877514839172, "perplexity_flag": "middle"}