f(x)=ax^2+bx+c with vertex given and one point need another point

Question

f(x)=ax^2 +bx +c      vertex is given as (1,8)  and the point (3,-4)  how do I find another point?

Answer 1

vertex form: f(x) =a(x-h)^2 +k

f(x) =a(x-1)^2 +8

find a by putting in (3,-4) for x, f(x)

-4=a(3-1)^2 +8

-4 =4a+8

4a=-12

a=-3

f(x) =-3(x-1)^2 +8

you can put in any number  for x and get f(x)

f(x) -3(0-1)^2 +8 = -3+8 =5

so (0,5) is another point

without finding the equation, since (1,8) is the vertex and x=1 the axis os symmetry , on the other side of the axis is the point (-1,-4)

Comments

for some reason: finding the points needed throws me -thankyou

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thanks a duplicate answer confirms what I had and what the other helper gave me