How do you find the intercepts for #y = x^2 +4x +3#?

1 Answer
Jul 17, 2018

Answer:

#x#-intercepts: #(-3, 0),(-1, 0)#

#y#-intercept: #(0, 3)#

Explanation:

Given: #y = x^2 + 4x + 3#

Find the x-intercepts by setting #y = 0#:

When the equation is in #Ax^2 + Bx + C# form, you can

factor the equation. Using the AC-Method :

Find two numbers #p " and " q# that multiply to #A*C = 3# and sum to #B = 4#:

#ul(" "p" "|" "q" "| " "p*q" "| " "p + q" ")#
#" "1" "|" "3" "| " "1*3 = 3" "|" "1 + 3 = 4 " WORKS!"#

#f(x) = (x + p)(x + q)#

#y = x^2 + 4x + 3 = (x +1)(x + 3)#

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

#x#-intercepts: #(-3, 0),(-1, 0)#

Find the y-intercept by setting #x = 0#:

y = 0^2 + 4*0 + 3 = 3

#y#-intercept: #(0, 3)#