Question

Using a directrix of y = 5 and a focus of (4, 1), what quadratic function is created?

Answer 1

f(x) = one fourth (x - 4)2 - 3

f(x) = one eighth (x + 4)2 - 3

f(x) = -one eighth (x - 4)2 + 3

f(x) = -one fourth (x + 4)2 - 3

Answer 2

The definition of a parabola is that any point on the parabola is the same distance from the focus and the directrix. Let (x,y) be a point on the parabola. Its distance from the focus (or distance squared) = (x - 4)^2 + (y -1)^2
The squared of the distance between the point (x,y) and y = 5 is (y - 5)^2

Equate the two distances (squared): (x - 4)^2 + (y -1)^2 = (y - 5)^2
Take the y term to RHS:
(x - 4)^2 = (y - 5)^2 - (y -1)^2
= (y - 5 + y - 1)(y - 5 - y + 1) (since a^2 - b^2 = (a+b)(a-b))
= (2y-6)(-4)
= 2(y-3)(-4)
=-8(y-3)
-1/8(x-4)^2 = y - 3

y = -1/8(x-4)^2 + 3

f(x) = -1/8(x-4)^2 + 3

Answer 3

Thank you again ranga, and if you could also help me with another?

Solve 2x^2 + 16x + 34 = 0

i got x= -4+i and x=-4-i
is that the same as (-4 +or- i)
or is it the same as (x-4 +or-i)
or is it neither?

Answer 4

2x^2 + 16x + 34 = 0
factor out 2
2(x^2 + 8x + 17) = 0
x^2 + 8x + 17 = 0
x = [ -8 +/- sqrt { (-8)^2 - (4)(1)(17) } ] / 2
= { -8 +/- sqrt(-4) } / 2
= -4 +/- i
Yes. -4 +/- i is shorthand representation of two solutions: -4 + i and -4 - i

Answer 5

thank you so much for your help and explanations!!!

Answer 6

You are very welcome.