How do you find the vertex and the intercepts for $y = x^2-5$ ?

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Answer 1 · 2018-06-05T14:28:04

vertex $=(0, -5)$

$"y-int" = 0$

$"x-int"=+-sqrt5$

Find the y-intercept set x=0:

$y = x^2-5=0^2-5=-5$

Find the x-intercept(s) if they exist set y=0:

$y = x^2-5$

$0 = x^2-5$

$5 = x^2$

$x=+-sqrt5$

axis of symmetry:

$ax^2+bx+c$

$y = x^2+0x-5$

$aos=-b/(2a) = -0/(2*1) = 0$

vertex #=(aos, f(aos))= (0, f(0))

$f(x) = x^2-5$

$f(0) = 0^2-5 = -5$

vertex $=(0, -5)$

graph{x^2 -5 [-10, 10, -5, 5]}

How do you find the vertex and the intercepts for y = x^2-5y = x^2-5?