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]}