Question

Find the equation of the parabola with vertex at (3,5) and focus at (3,-2)?

Answer 1

a=-7, so in vertex form, (x-h)^2=4(a)(y-k) gives
(x-3)^2=4(-7)(y-5) or (x-3)^2=-28(y-5)
if you want this in y= form, just solve for y.

Answer 2

Thanks.
so y= (-1/28)( x-3)^2
or y = -1/28(x^2 -6x +9)

Answer 3

i got y = (-1/28)(x-3)^2 + 5

Answer 4

3. y = x + 3
a) y-intercept:
set x=0
re-write the equation: y=x+3 : set x=0
final equation: y=3
hence y-intercept =3
b x-intercep:
set y=0
re-write the equation y=x+3 and set y= 0
x+3 =0
x =-3
x-intercept =-3
graph is as follows with the intercepts


4. y = -2x-7
a) y-intercept:
set x=0
re-write the equation: y=0*(-2) -7
final equation: y= -7
hence y-intercept = -7
b) y = -2x-7
for x-intercept:
set y=0
re-write the equation:
-2x -7 =0
-2x = 7
x = -7/2
x-intercept =-7/2
graph is as follows with the intercepts






5. 2x + 3y =9
a) y-intercept:
set x=0
re-write the equation: 2*0 +3y =9
final equation: 3y =9
hence y-intercept = 3
b) 2x + 3y =9
for x-intercept:
set y=0
re-write the equation:
2x + 3y =9 = 2x + 3*0 =9
2x =9
x = 9/2
x-intercept = 9/2 = 4.5
graph is as follows with the intercepts