Question

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

Answer 1

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

Answer 2

(a)
`y=x^2-2x-8`

(b)
`y= -\frac12x^2+4x-6`

Explanation:

(a) From inspection of the graph, the x-intercepts (where
`y = 0`) are at

`x = -2` and
`x =4`.

Therefore
`y = a(x + 2)(x - 4)` where
`a` is some constant

The y-intercept is at (0, -8)

Expanding the equation:

`y = a(x + 2)(x - 4)`

`\implies y = a(x^2-4x+2x-8)`

`\implies y= ax^2-2ax-8a`

Therefore, -8a is the y-intercept

Comparing y-intercepts: -8 = -8a ⇒ a = 1

Substituting a = 1, the final equation in standard form is:

`y=x^2-2x-8`

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(b) From inspection of the graph, the x-intercepts (where
`y = 0`) are at

`x = 2` and
`x =6`.

Therefore
`y = a(x -2)(x - 6)` where
`a` is some constant

The y-intercept is at (0, -6)

Expanding the equation:

`y = a(x -2)(x - 6)`

`\implies y = a(x^2-6x-2x+12)`

`\implies y= ax^2-8ax+12a`

Therefore, 12a is the y-intercept

Comparing y-intercepts:

`-6 = 12a \implies a = -\frac12`

Substituting found value of a into the equation:

`\implies y= (-\frac12)x^2-8(-\frac12)x+12(-\frac12)`

`\implies y= -\frac12x^2+4x-6`