# Does the function f(x)= -x^2+6x-1 have a minimum or maximum value?

Jan 28, 2017

The parabola will have a maximum value because the ${x}^{2}$ term is negative.

#### Explanation:

1. Because the function has the general form $f \left(x\right) = \textcolor{b l u e}{A} {x}^{2} + \textcolor{p u r p \le}{B} x + \textcolor{red}{C}$, we know the graph will be a parabola.
2. The sign of the ${x}^{2}$ term will tell us if the parabola opens up (like a $\cup$) or down (like a $\cap$):

If $A > 0$, opens up ($\cup$)

If $A < 0$, opens down ($\cap$)

In this case,
$f \left(x\right) = \textcolor{b l u e}{- \left(1\right)} {x}^{2} \textcolor{p u r p \le}{+ 6} x \textcolor{red}{- 1}$
$\textcolor{b l u e}{A = - 1}$ so the parabola will open "down" or $\cap$ which means the parabola will have a "peak" or maximum point.

graph{-x^2+6x-1 [-15, 15, -10, 10]}
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}