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

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Answer 1 · 2015-12-21T16:42:04

$x$ $x$ -intercept @ $x = - 2$ $x=-2$
$y$ $y$ -intercept @ $y = 4$ $y=4$

Original Equation:
$y = x^{2} + 4 x + 4$ $y=x^2+4x+4$

$x$ $x$ -intercept
Factor:
$y = (x + 2) (x + 2)$ $y=(x+2)(x+2)$

Solve:
$x = - 2$ $x=-2$

$x$ $x$ -intercept @ $x = - 2$ $x=-2$

$y$ $y$ -intercept
Set $x$ $x$ equal to zero and solve:
$y = {(0)}^{2} + 4 (0) + 4$ $y=(0)^2+4(0)+4$
$y = 0 + 4$ $y=0+4$
$y = 4$ $y=4$

$y$ $y$ -intercept @ $y = 4$ $y=4$

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