How do you graph #f(x)=(x-2)^2# and identify the x intercepts, vertex?

1 Answer
Jul 13, 2017

#"see explanation"#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#
where ( h , k ) are the coordinates of the vertex and a is a constant.

#y=(x-2)^2" is in this form"#

#"with " a=1,h=2,k=0#

#rArrcolor(magenta)"vertex "=(2,0)#

#"since this point is on the x-axis it is also the x-intercept"#

#color(blue)"shape of graph"#

#• " if "a>0" then minimum turning point " uuu#

#• " if "a<0" then maximum turning point " nnn#

#"here " a=1>0rArr" minimum turning point"#

#color(blue)"some other points"#

#x=0toy=(0-2)^2=4larrcolor(red)" y-intercept"#

#x=-1toy=(-3)^2=9rArr(-1,9)#

#x=3toy=1rArr(3,1)#

#"plot these points and draw a smooth curve through them"#
graph{(x-2)^2 [-10, 10, -5, 5]}