How do you find the vertex and intercepts for #y = (x + 3)^2 – 4#?

3 Answers
Dec 29, 2017

Vertex #->(x,y)=(-3,-4)#
#x_("intercepts")->(x,y) = (-5,0) and (-1,0)#
#y_("intercept")->(x,y)=(0, 5)#

Explanation:

#color(blue)("Determine the "x" intercepts")#

Set # y=0=(x+3)^2-4#

Add 4 to both sides

#4=(x+3)^2#

Square root both sides

#+-2=x+3#

Subtract 3 from both sides

#x=-3+-2#

#x_("intercepts")->(x,y) = (-5,0) and (-1,0)#

You may if you so choose determine the vertex from this point

#x_("vertex")# is midway between the x intercepts then by substitution determine y. The method I used later takes less work.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determin "y" intercept and vertex")#

Given #y(xcolor(magenta)(+3))^2color(green)(-4)#

#x_("vertex")=(-1)xx(color(magenta)(+3))=-3#
#y_("vertex")=color(green)(-4)#

Vertex #->(x,y)=(-3,-4)#

#y_("intercept")=color(magenta)(+3)^2color(green)(-4) = 5#

Tony B

Dec 29, 2017

Vertex is at #(-3, -4) # , y intercept is #y=5 or (0,5)#
x intercepts are at #(-1,0) and (-5,0)#

Explanation:

#y=(x+3)^2-4#

Comparing with vertex form of equation

#f(x) = a(x-h)^2+k ; (h,k)# being vertex we find

here #h=-3 , k=-4 :.# Vertex is at #(-3, -4) #

y intercept is found by putting #x=0# in the equation

#y = (x+3)^2-4 :. y=(0+3)^2-4 =9-4=5#

So y intercept is #y=5 or (0,5)#

x intercepts are found by putting #y=0# in the equation

#y = (x+3)^2-4 :. 0=(x+3)^2-4 # or

#(x+3)^2=4 or (x+3) =+-sqrt4 or (x+3) =+-2#

#:. x=-3+-2 :. x=-1 and x=-5#

x intercepts are at #(-1,0) and (-5,0)#

graph{(x+3)^2-4 [-10, 10, -5, 5]} [Ans]

Dec 29, 2017

Vertex = #(-3,-4)#

y-intercept = #(0,5)#

x-intercepts = #(-1,0)# and #(-5,0)#

Explanation:

First, expand #y=(x+3)^2-4# into #y=x^2+6x+5#.

The x-vertex of a quadratic equation #y=ax^2+bx+c# is given by #-b/2a#.

Plugging in #a=1#, #b=6#, and #c=5#, we find that the x-vertex is #-6/2=-3#.

Now we plug in #x=-3# into #y=x^2+6x+5# to get the y-vertex, which is #-4#.

So our vertex is #(-3,-4)#.

To find the y-intercept of a quadratic equation, plug in #0# for the x value, and you will get #5#.

To find the x-intercept of a quadratic equation, plug in #0# for the y value, and you will get two solutions, #x=-1# and #x=-5#.

So our y-intercept is #(0,5)#, and our x-intercepts are #(-1,0)# and #(-5,0)#.