How do you write a quadratic equation with x -intercepts: -4,1 and point(: ( -3, -4)?

1 Answer
Mar 3, 2017

Quadratic equation is #y=x^2+3x-4#

Explanation:

Le the quadratic equation be #y=ax^2+bx+c#

as #x#-intercepts are #-4# and #1#, we have

#a(-4)^2+b(-4)+c=0# or #16a-4b+c=0# ...........(1)

and #a+b+c=0# ...........(2)

As it passes through #(-3,-4)#, we have

#-4=a(-3)^2+b(-3)+c# or #9a-3b+c=-4# ...........(3)

Subtracting (2) from (1), we get #15a-5b=0# or #3a-b=0# ........(4)

and subtracting (3) from (1), we get #7a-b=4# ...........(5)

Subtracting (4) from (5), we get #4a=4# i.e. #a=1#

Putting this in (4), we get #3-b=0# or #b=3#

and putting #a# and #b# in (2), we get #c=-4#

Hence quadratic equation is #y=x^2+3x-4#

graph{y=x^2+3x-4 [-12.21, 7.79, -6.48, 3.52]}