How do you find the intercept and vertex of $y=(-x-1)(x+7)$ ?

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Answer 1 · 2018-02-12T06:26:08

X-intercepts

$(-1, 0)$
$(-7, 0)$
Y- intercept
$(0, -7)$
Vertex
$(-4. 9)$

$y=(-x-1)(x+7)$
$y=-x^2-x-7x-7$
$y=-x^2-8x-7$

X-intercepts
At $y=0$
$(-x-1)(x+7)=0$
$-x-1=0$
$-x=1$
$x=-1$
$(1, 0)$
$x+7=0$
$x=-7$
$(-7, 0)$

Y- intercept
At $x=0$

$y=(0-1)(0+7)$
$y=-7$
$(0, -7)$

Vertex

$y=-x^2-8x-7$
$x=(-b)/(2a)=8/(-2)=-4$
At $x=-4$
$y=-(-4)^2-8(-4)-7$
$y=-16+32-7=-23+32=9$
$(-4. 9)$

How do you find the intercept and vertex of y=(-x-1)(x+7)y=(-x-1)(x+7)?