Question

Help asap!

Thanks in advance!

There's two questions btw:)

Answer 1

[1].

Option A and D are correct.

[2].

Option A is correct

Explanation:

[1].

Quadratic function states that it is an equation of second degree i.,e it contains at least one term that is squared.

The standard form of the quadratic equation is;
`ax^2+bx+c = 0`

A.

`y(y+4)-y = 6`

Using distributive property:
`a\cdot (b+c) = a\cdot b + a\cdot c`

`y^2+4y-y=6`

Combine like terms;

`y^2+3y = 6`

or

`y^2+3y -6=0` which represents a quadratic equation.

B.

`3a-7 = 2(7a-3)`

`3a-7 = 14a-6`

or

`11a+1 = 0` which is not a quadratic equation.

C.

(3x+2)+(6x-1) = 0

Combine like terms;

9x +1 = 0 which is not a quadratic equation.

D.

4b(b) = 0

`4b^2 = 0` which represents the quadratic equation.

[2].

Given the parent function:
`y=x^2`

The reflection rule over x axis is given by;

`(x, y) \rightarrow (x, -y)`

then

the function become:
`y = -x^2`

Vertical shift:

If c is a positive real number, the graph y=f(x)+c is the graph of y =f(x) shifted upward c units.

If c is a positive real number, the graph y=f(x)-c is the graph of y =f(x) shifted downward c units.

then;

The graph
`y=-x^2-3` is the graph of
`y=-x^2` shifted 3 units down.

Therefore, the translation of the graph of
`y=x^2` to obtain
`y=-x^2-3` is, reflect over the x-axis and shift down 3