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

Jul 17, 2018

$x$-intercepts: $\left(- 3 , 0\right) , \left(- 1 , 0\right)$

$y$-intercept: $\left(0 , 3\right)$

#### Explanation:

Given: $y = {x}^{2} + 4 x + 3$

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

When the equation is in $A {x}^{2} + B x + C$ form, you can

factor the equation. Using the AC-Method :

Find two numbers $p \text{ and } q$ that multiply to $A \cdot C = 3$ and sum to $B = 4$:

$\underline{\text{ "p" "|" "q" "| " "p*q" "| " "p + q" }}$
$\text{ "1" "|" "3" "| " "1*3 = 3" "|" "1 + 3 = 4 " WORKS!}$

$f \left(x\right) = \left(x + p\right) \left(x + q\right)$

$y = {x}^{2} + 4 x + 3 = \left(x + 1\right) \left(x + 3\right)$

$\left(x + 1\right) \left(x + 3\right) = 0 \implies x = - 1 , x = - 3$

$x$-intercepts: $\left(- 3 , 0\right) , \left(- 1 , 0\right)$

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

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

$y$-intercept: $\left(0 , 3\right)$