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

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Answer 1 · 2018-07-17T03:47:24

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

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

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

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

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

factor the equation. Using the AC-Method :

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

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

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

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

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

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

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

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

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

How do you find the intercepts for y = x^2 +4x +3y=x2+4x+3y = x^2 +4x +3?