Question

Help with these questions please

Answer 1

equation for parabola with x as quadratic:
\[y-k = 4a(x-h)^2\]

Answer 2

Yes

Answer 3

where (h,k) are the coordinates of the vertex

Answer 4

Ok

Answer 5

rearranging you'll have
\[y - k = 4a (x-h)^2\]
substituting the values h and k
\[y - (-3) = 4a(x - 5)^2\]
rearrange/transpose such that it would look like this
\[y = 4a( x-h)^2 + k \]

Answer 6

Is that the answer?

Answer 7

not yet. we still have to check the equations if it will pass through the point (6,1)

Answer 8

oh o ok

Answer 9

\[y - (-3) = 4a(x-5)^2\]
\[y + 3 = 4a(x-5)^2\]
\[y = 4a(x-5)^2 - 3\]
in the choices given you have
\[y = 4(x-5)^2 - 3\]
check whether (6,1) passes through this curve
substitute x=6 to the equation \[y = 4(x-5)^2 - 3\]

Answer 10

ohhh

Answer 11

How about the other one?

Answer 12

\[y = 4(x-5)^2 - 3\]
\[y = 4(6-5)^2 - 3\]
\[y = 4(1) - 3\]
\[y = 1\]
since this corresponds to the value of y in the point (6,1) then the line passes thru the point given

Answer 13

by inspection, what would be the coordinates of the vertex of the parabola?

Answer 14

5,-3

Answer 15

the one that's being graphed?

Answer 16

noo

Answer 17

i'm talking of the second one

Answer 18

oh ok

Answer 19

0,150

Answer 20

u sure?

Answer 21

I think..

Answer 22

we're talking of the vertex here. in the graph it's the peak of the curve

Answer 23

50,150

Answer 24

yep

Answer 25

Really??

Answer 26

Hello??

Answer 27

yep. and since we know that the curve passes thru the point (0,0) then we'll have
\[y - 150 = 4a(x-50)^2\]
\[y = 4a(x-50)^2 + 150\]
substituting (0,0)
\[0 = 4a(0-50)^2 + 150\]
\[4a(50^2) = - 150\]
\[a = \frac{ -150 }{ 4(50^2) }\]
\[a = - \frac{ 3 }{ 200 }\]
substituting to the equation you'll have
\[y = 4a(x-50)^2 + 150\]
\[y = 4(-\frac{ 3 }{ 200 })(x-50)^2 + 150\]
\[y = - \frac{ 3 }{ 50 }(x-50)^2 + 150\]