Quite urgent lol but should probably be quick I guess :(

Answer:
(a) [tex]y=x^2-2x-8[/tex]
(b) [tex]y= -\dfrac12x^2+4x-6[/tex]
Step-by-step explanation:
(a) From inspection of the graph, the x-intercepts (where [tex]y = 0[/tex]) are at
[tex]x = -2[/tex] and [tex]x =4[/tex].
Therefore [tex]y = a(x + 2)(x - 4)[/tex] where [tex]a[/tex] is some constant
The y-intercept is at (0, -8)
Expanding the equation:
[tex]y = a(x + 2)(x - 4)[/tex]
[tex]\implies y = a(x^2-4x+2x-8)[/tex]
[tex]\implies y= ax^2-2ax-8a[/tex]
Therefore, -8a is the y-intercept
Comparing y-intercepts: -8 = -8a ⇒ a = 1
Substituting a = 1, the final equation in standard form is:
[tex]y=x^2-2x-8[/tex]
--------------------------------------------------------------------------------------------------
(b) From inspection of the graph, the x-intercepts (where [tex]y = 0[/tex]) are at
[tex]x = 2[/tex] and [tex]x =6[/tex].
Therefore [tex]y = a(x -2)(x - 6)[/tex] where [tex]a[/tex] is some constant
The y-intercept is at (0, -6)
Expanding the equation:
[tex]y = a(x -2)(x - 6)[/tex]
[tex]\implies y = a(x^2-6x-2x+12)[/tex]
[tex]\implies y= ax^2-8ax+12a[/tex]
Therefore, 12a is the y-intercept
Comparing y-intercepts:
[tex]-6 = 12a \implies a = -\dfrac12[/tex]
Substituting found value of a into the equation:
[tex]\implies y= (-\dfrac12)x^2-8(-\dfrac12)x+12(-\dfrac12)[/tex]
[tex]\implies y= -\dfrac12x^2+4x-6[/tex]
y = a(x-h)²+k where (h,k) denotes the vertex
Q.1)
vertex: (1, -9) where h=1, k = -9
coordinates: (0, -8) where x = 0, y = -8
find for a:
y = a(x-h)²+k
-8 = a(0-1)²-9
-8 = a -9
a = 1
============
equation: y = (x-1)²-9
Q.2)
vertex: (4, 2) where h=4, k = 2
coordinates: (2, 0) where x = 2, y = 0
find for a:
y = a(x-h)²+k
0 = a(2-4)²+2
-2 = 4(a)
a = -1/2
============
equation: y = -1/2(x-4)²+2