How do you graph the parabola #y=2(x+1)^2+3# using vertex, intercepts and additional points?

1 Answer
Jun 2, 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=2(x+1)^2+3" is in this form"#

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

#rArrcolor(magenta)"vertex "=(-1,3)#

#color(blue)"Intercepts"#

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

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

#x=0toy=2(0+1)^2+3=5larrcolor(red)" y-intercept"#

#y=0to2(x+1)^2+3=0#

#rArr(x+1)^2=-3/2#

#"but " (x+1)^2>=0#

#rArr" there are no x-intercepts"#

#color(blue)"Shape of parabola"#

#• " if " a>0" then minimum " uuu#

#• " if " a<0" then maximum " nnn#

#"here " a=2rArr" minimum turning point"#
graph{2(x+1)^2+3 [-10, 10, -5, 5]}