How do you find the vertex and the intercepts for #y=(x+3)(x-1)#?

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Answer 1 · 2016-08-30T19:12:53

Vertex: #(-1, -4)#
x-intercepts: #(-3, 0)# and #(1, 0)#
y-intercept: #(0, -3)#

Since the function is already factored, we can note the x-intercepts.

#0 = (x+ 3)(x - 1)#

#x = -3 and 1#

Hence, the x-intercepts are at #(-3, 0)# and #(1, 0)#.

We should now multiply out.

#y = (x + 3)(x - 1)#

#y =x^2 + 3x - x - 3#

#y = x^2 + 2x - 3#

We can now find the y-intercept very easily.

#y = 0^2 + 2(0) - 3#

#y = -3#

Hence, the y-intercept is at #(0, -3)#.

As for the vertex, it'll be easiest to complete the square to convert into vertex form, #y = a(x - p)^2 + q#.

#y = x^2 + 2x - 3#

#y = 1(x^2 + 2x + n - n) - 3#

#n = (b/2)^2 = (2/2)^2 = 1#

#y = 1(x^2 + 2x + 1 - 1) - 3#

#y= 1(x^2 + 2x + 1) - 1 - 3#

#y = 1(x + 1)^2 - 4#

The vertex in vertex form #y = a(x - p)^2 + q# is located at #(p, q)#. Thus, our vertex is at #(-1, -4)#.

Hopefully this helps!