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

Question

Isaiah Anthony · Accepted Answer

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!

Jace Anderson · Answer

undefined To find the vertex of the quadratic function $ y = (x+3)(x-1) $, you first need to expand the expression to obtain the quadratic in standard form $ y = ax^2 + bx + c $. Then, you can find the x-coordinate of the vertex using the formula $ x = -\frac{b}{2a} $. To find the y-coordinate of the vertex, substitute the x-coordinate you found back into the original function.

To find the x-intercepts, set $ y = 0 $ and solve for x.

To find the y-intercept, set $ x = 0 $ and evaluate the function.