How do you graph the parabola #y=(x-2)^2-3# using vertex, intercepts and additional points?

1 Answer
Jul 3, 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)^2-3" is in this form"#

#"with " h=2,k=-3" and " a=1#

#rArrcolor(magenta)"vertex "=(2.-3)#

#color(blue)"shape of parabola"#

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

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

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

#color(blue)"Intercepts"#

#• " let x = 0, in equation for y-intercept"#

#• " let y = 0, in equation for x-intercepts"#

#x=0toy=(0-2)^2-3=1larrcolor(red)" y-intercept"#

#y=0to(x-2)^2-3=0#

#rArr(x-2)^2=3larr" take square root of both sides"#

#rArrx-2=+-sqrt3larr" note plus or minus"#

#rArrx=2+-sqrt3#

#x~~0.27,x~~3.73larrcolor(red)" x-intercepts"#

#color(blue)"Additional points"#

#"choose values for x and evaluate for y"#

#"Example"#

#x=-1toy=9-3=6rArr(-1,6)#

#x=4toy=4-3=1rArr(4,1)#

#"plot vertex, intercepts, additional points and draw a "#
#"smooth curve through them"#
graph{(x-2)^2-3 [-10, 10, -5, 5]}