If h(x) = 25-x^2
or f(x) = (x-6)^2+3
I know finding vertex is -b/2a But for those, it doesn't really give b or a I believe.. How would i go about solving this? Thanks!
or f(x) = (x-6)^2+3
I know finding vertex is -b/2a But for those, it doesn't really give b or a I believe.. How would i go about solving this? Thanks!
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rewrite it:
-(x^2-25)
this is difference of squares so now we have:
-(x-5)(x+5) the x intercepts are 5, -5
For vertex: find the axis of symmetry from the x-intercept: x=0, plug it in into the original equation and we have y=25 so thats the vertex
Same thing for the second problem:
rewrite it:
x^2 -12x +39
now we can have and a and b, so the vertex is (6,3)
-(x^2-25)
this is difference of squares so now we have:
-(x-5)(x+5) the x intercepts are 5, -5
For vertex: find the axis of symmetry from the x-intercept: x=0, plug it in into the original equation and we have y=25 so thats the vertex
Same thing for the second problem:
rewrite it:
x^2 -12x +39
now we can have and a and b, so the vertex is (6,3)
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Math isnt real!!!! Its a conspiracy!