How do you find the vertex and intercepts for #y = (x + 5)^2 – 2#?

Question

1039 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-19T10:00:04

Vertex->(x,y)=(-5,-2)#

#y_("intercept")->y=23#

#x_("intercept")# as follows:
#x=-5+-sqrt(2)#

#x~~-6.414# to 3 decimal places
#x~~-3.586# to 3 decimal places

#color(blue)("Determine the vertex")#

This is already in a vertex form the of quadratic equation so you can read off the coordinates of the vertex directly.

#x_("vertex")=(-1)xx(+5) = -5#
#y_("vertex")=-2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine y intercept")#

y intercept is at #x=0#

#=>y=(0+5)^2-2 -> y=23#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine x intercept")#

Set #y=0# giving

#0=(x+5)^2-2#

Add 2 to both sides

#0+2=(x+5)^2-2+2#

#=>2=(x+5)^2" "vec("swap round") " " (x+5)^2=2#

Square root both sides

#x+5=sqrt(2)#

Subtract 5 from both sides

#x=-5+-sqrt(2)#

#x~~-6.414# to 3 decimal places
#x~~-3.586# to 3 decimal places

Tony B