Question

how to find the focus of y =(1/8)x^2 -x +3

Answer 1

Well RULES FIRST:
lets say that equation is x=y^2 + 4,
in this one i know that focus is (6,0) but how do you obtain it?

1) Converting this into standard form, it is, "y^2 = x - 4"

2) A standard form of parabola equaiton is "(y - k)^2 = 4a(x - h)'
whose, vertex is (h. k) and focus is (a + h, k)

3) Comparing the equation in (1) with the standard form in (2),
h = 4, k = 0 and a = 1/4

4) Hence Focus is (4 + 1/4, 0) = (17/4, 0)

You may kindly verify how it is (6, 0)

Answer 2

\[
y=\frac{x^2}{8}-x+3=\frac{1}{8}
\left(x^2-8
x\right)+3=\\\frac {1}{8}
\left((x-4)^2-16\right)+3=\frac{1}{8} (x-4)^2+1
\]

Answer 3

\[
y-1 =\frac{1}{8} (x-4)^2
\]

Answer 4

thanks so much!

Answer 5

Sorry Focus is
(4,1+2)=(4,3)

Answer 6

See the attached graph.