Question 238868
What is the coordinates of the turning point of the graph of the function y=x^2-6x+5.
<pre><font size = 4 color = "indigo"><b>
The turning point of {{{y = ax^2+bx+c}}} is {{{"("}}}{{{-b/2}}}{{{","}}}{{{-d/4}}}{{{")"}}}

where {{{d=b^2-4ac}}}

For {{{y = 1x^2-6x+5}}}, {{{a=1}}}, {{{b=-6}}}, {{{c=5}}}

{{{d=b^2-4ac=(-6)^2-4(1)(5)=36-20=16}}}

Turning point = {{{"("}}}{{{-b/2}}}{{{","}}}{{{-d/4}}}{{{")"}}}{{{"="}}}{{{"("}}}{{{"("}}}{{{-(-6)/2}}}{{{","}}}{{{-(16)/4}}}{{{")"}}}{{{"="}}}{{{"("}}}{{{-(-3)}}}{{{","}}}{{{-4}}}{{{")"}}}{{{"="}}}{{{"("}}}{{{3}}}{{{","}}}{{{-4}}}{{{")"}}}{{{"="}}}

{{{drawing(285.7,400,-2,8,-6,8, green(line(3,0,3,-4)), green(line(0,-4,3,-4)),

graph(285.7,400,-2,8,-6,8,x^2-6x+5), locate(3,-4,"(3,-4)") )}}}

Edwin</pre>