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

Mar 3, 2017

Quadratic equation is $y = {x}^{2} + 3 x - 4$

#### Explanation:

Le the quadratic equation be $y = a {x}^{2} + b x + c$

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

$a {\left(- 4\right)}^{2} + b \left(- 4\right) + c = 0$ or $16 a - 4 b + c = 0$ ...........(1)

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

As it passes through $\left(- 3 , - 4\right)$, we have

$- 4 = a {\left(- 3\right)}^{2} + b \left(- 3\right) + c$ or $9 a - 3 b + c = - 4$ ...........(3)

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

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

Subtracting (4) from (5), we get $4 a = 4$ 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 $y = {x}^{2} + 3 x - 4$

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