Question

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

Answer 1

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]}