How do you find the vertex and the intercepts for #f(x) =(x+5)^2 -1#?

1 Answer
Nov 14, 2017

#(-5,-1)" and "x=-6,x=-4#

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 multiplier"#

#f(x)=(x+5)^2-1" is in vertex form"#

#"with "h=-5" and "k=-1#

#rArrcolor(magenta)"vertex "=(-5,-1)#

#color(blue)"Intercepts"#

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

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

#x=0toy=(5)^2-1=24larrcolor(red)"y-intercept"#

#y=0to(x+5)^2-1=0#

#rArr(x+5)^2=1#

#color(blue)"take the square root of both sides"#

#sqrt((x+5)^2)=+-sqrt1larrcolor(blue)"note plus or minus"#

#rArrx+5=+-1#

#"subtract 5 from both sides"#

#rArrx=-5+-1#

#x=-5-1=-6,x=-5+1=-4larrcolor(red)"x-intercepts"#