How do you find the vertex and the intercepts for $y=x^2 - 9x$ ?

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Answer 1 · 2017-06-28T06:23:51

Vertex $(4.5, -20.25)$
Intercepts $(0,0); (0,9)$

Given -

$y=x^2-9x$

Vertex

$x=(-b)/(2a)= (-(-9))/(2xx1)=9/2=4.5$
At $x=4.5$
$y=(4.5)^2-9(4.5)$
$y=20.25-40.5=-20.25$

Vertex $(4.5, -20.25)$

Intercepts
At $y=0$

$x^2-9x=0$
$x(x-9)=0$
$x=0$
$x=9$

Intercepts $(0,0); (0,9)$

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How do you find the vertex and the intercepts for y=x^2 - 9xy=x^2 - 9x?