Question

General form of a parabola
vertex- (4,1)
x-intercepts (3,0) and (5,0)

Answer 1

ok... do you know the vertex form for the equation for a parabola???

Answer 2

y=a(x-h)^2+k

Answer 3

ok... good so we have:
\(\large y=a(x-4)^2+1 \)

since we know the parabola passes through the given points, plug those values in so we can find "a"....

Answer 4

All I know is that the vertex is (h,k) so y=a(x-4)^2+1
I just don't know how to find a

Answer 5

i'm a mind reader!!!

Answer 6

Lol

Answer 7

ok... let's plug in (3,0):
\(\large 0=a(3-4)^2+1 \)

solve for a....
wat u got???

Answer 8

0=0?

Answer 9

not quite... try again...

Answer 10

\(\large 0=a(3-4)^2+1 \)
\(\large 0=a(-1)^2+1 \)
\(\large 0=a\cdot 1+1 \)
\(\large 0=a+1 \)

a = ???

Answer 11

Omg, I'm retarded lol
-1=a

Answer 12

ok... just to make sure, let's see if we get the same value for a if we plugged in the other point (5, 0)...

do we get the same value???

Answer 13

yep :)

Answer 14

good, so the VERTEX form is: \(\large y=-(x-4)^2+1 \)
you want GENERAL form so what we need to do is expand the right side and move the y over to the right side.....

Answer 15

So the equation would be y=-1(x-4)^2+1

Answer 16

mind reader!!!

Answer 17

Lol, it's perfect because your owl looks shocked

Answer 18

haha...:) ok... expand that right side....

Answer 19

0=-y-(x-4)^2+1

Answer 20

ok.. general form looks like this: \(\large 0=Ax^2+Bx+Cy+D \)
so you need to expand (x-4)^2

Answer 21

(x-4)^2=x^2-8x+16

Answer 22

ok... so we still have:
\(\large 0= -y-(x^2-8x+16)+1\)
so arranging and simplifying them we have:
\(\large 0= -y-x^2+8x-16+1\)
\(\large 0= -x^2+8x-y-15\)

OR

\(\large x^2-8x+y+15=0\)

Answer 23

So that's the answer?
x^2-8x+y+15=0

Answer 24

either answer will work... they're both in general form....

Answer 25

yes... :)

Answer 26

Thank you so much!

Answer 27

yw....:)