Question

Help asap! Thanks in advance! There's two questions btw:)

Answer 1

Answer:

[1].

Option A and D are correct.

[2].

Option A is correct

Step-by-step 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