How do you find the x intercepts and vertex of $y=x^2-2x-4$ ?

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Answer 1 · 2016-12-15T02:20:54

X-intercepts are

$(1-sqrt5, 0)$
$(1+sqrt5, 0)$

Vertex $(1, -5)$

Given -

$y=x^2-2x-4$

To find the x-intercepts put $y=0$

$x^2-2x-4=0$
$x^2-2x=4$
$x^2-2x+1=4+1$
$(x-1)^2=5$
$x-1=+-sqrt5$

X-intercepts are

$(1-sqrt5, 0)$
$(1+sqrt5, 0)$

Vertex

$x=(-b)/(2a)=(-(-2))/(2xx1)=2/2=1$

At $x=1$

$y=(1)^2-2(1)-4=1-2-4=-5$

Vertex $(1, -5)$

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How do you find the x intercepts and vertex of y=x^2-2x-4y=x^2-2x-4?