# Thread:

1. ## Reciprocal of a Quadratic Function

Say you have a function f(x)=1/(x^2+6x+11). Would you have to complete the square to figure out the vertex of the parabola in the denominator(then sub in the x-value in the function to figure out the vertex in the rational function), or is there an easier way to draw the graph?

2. ## Re: Reciprocal of a Quadratic Function

Personally, I've never heard of the vertex of a denominator? ...
If you have to find the vertex of a function (in this case a rational one) then the first derivative can be a good option.

3. ## Re: Reciprocal of a Quadratic Function

Originally Posted by Dragon08
Say you have a function f(x)=1/(x^2+6x+11). Would you have to complete the square to figure out the vertex of the parabola in the denominator(then sub in the x-value in the function to figure out the vertex in the rational function), or is there an easier way to draw the graph?
The quadratic function in the denominator is positive for all x. The vertex of the quadratic function is an absolute minimum for that function. The minimum of the quadratic function in the denominator, corresponds to the absolution maximum of the rational function, in that they both have the same x coordinate. The maximum of the rational function is equal to the reciprocal of the minimum for the quadratic function.

4. ## Re: Reciprocal of a Quadratic Function

Originally Posted by SammyS
The quadratic function in the denominator is positive for all x. The vertex of the quadratic function is an absolute minimum for that function. The minimum of the quadratic function in the denominator, corresponds to the absolution maximum of the rational function, in that they both have the same x coordinate. The maximum of the rational function is equal to the reciprocal of the minimum for the quadratic function.
So you still have to figure out the minimum value of the quadratic function by completing the square right?

5. ## Re: Reciprocal of a Quadratic Function

Yes, and for this problem that is very simple. 6/3= 3 and $3^2= 9$. 11= 9+ 3

6. ## Re: Reciprocal of a Quadratic Function

Originally Posted by HallsofIvy
Yes, and for this problem that is very simple. 6/3= 3 and $3^2= 9$. 11= 9+ 3

Someone needs to check their work!

7. ## Re: Reciprocal of a Quadratic Function

See the graph. The vertex is at ( $-3,2$).

This could also have been calculated using differentiation.

Since, the parabola is of the form $X^2=4aY$, the slope at the vertex must be equal to zero.

Therefore, $y=x^2+6x+11$ <=> $\frac{dy}{dx}=2x+6$<=> $0=2x+6$<=> $x=-3$
Attached Thumbnails
{"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": 8, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8952240347862244, "perplexity_flag": "head"}