Question

Suppose a parabola has vertex (-8,-7) and also passes through the point (-7,-4). What is the equation of the parabola in vertex form

Answer 1

just spit-balling here, but you could try simultaneous equations for this:

y = ax^2 + bx + c
-4 = 49a -7b + c ENQ 1

also
-7 = 64a - 8b +c EQN 2

and
y' = 2ax + b
0 = -16a + b EQN 3

Answer 2

... so b = 16a

subbed into EQN 2
-7 = 64a - 8b +c
-7 = 64a - 8(16a) +c
-7 = 64a - 128a + c
-7 = -64a + c
so c = -7 + 64a

Answer 3

@Jack1 you used information for y' , where did you get that?

Answer 4

subbed into Equation 1
-4 = 49a -7b + c ENQ 1
-4 = 49a -7b -7 + 64a
-4 = 49a -7(16a) -7 + 64a
-4 = 49a -112a -7 + 64a
-4 = a -7
so a = 3

so
0 = -16a + b
0 = -48 + b
b = 48

Answer 5

c = -7 + 64a
c = -7 + 64*3
c = -7 + 192
c = 185

Answer 6

@mathessentials y' = derivative of the line
at the vertex, y' = turning point so gradient = 0

Answer 7

I see :) it makes sense, thanks for explaining

Answer 8

all good dude

Answer 9

anyways, i think i went the wrong way, this gives you the equation of the line in the form
\[y = ax^2 + bx + c\] in this case - \[y = 3x^2 + 48x + 185\]
whereas I think vertex form is something like:
\[y = a(x – h)^2 + k\] where "k" is ... the y value of the vertex?
and "h" = the x value?
I know "a" is the same in either...

so your vertex form equation should look something more like \[y = 3(x + 8)^2 -7\] ... maybe...?