For a function to be even, it must satisfy the equality \(\displaystyle f(x)=f(-x)\)

Likewise if a function is even, it is symmetrical about the y-axis \(\displaystyle f(-x)=(-x)^3+5(-x)-7=-x^3-5x-7\neq f(x)\)

Therefore, the function is not even, and so the answer is No

For a function to be symmetrical about the y-axis, it must satisfy \(\displaystyle f(x)=f(-x)\) \(\displaystyle f(-x)=\sqrt{4+x}\neq f(x)\) so there is not symmetry about the y-axis

For a function to be symmetrical about the x-axis, it must satisfy \(\displaystyle f(-x)=-f(x)\) \(\displaystyle f(-x)=\sqrt{4+x}=-f(x)\) so there is symmetry about the x-axis

For a function to be symmetrical about the origin, you must replace y with (-y) and x with (-x) and the resulting function must be equal to the original function.

\(\displaystyle (-x)=-(-y)^2+4\neq x=-y^2+4\)

-So there is no symmetry about the origin, and the answer is Symmetrical about the x-axis

For a function to be symmetrical about the y-axis, it must satisfy \(\displaystyle f(x)=f(-x) f(-x)=|-x|+2=|x|+2\) so there is symmetry about the y-axis

For a function to be symmetrical about the x-axis, it must satisfy \(\displaystyle f(-x)=-f(x)\) 

\(\displaystyle f(-x)=|-x|+2\neq -f(x)\) so there is not symmetry about the x-axis

For a function to be symmetrical about the origin, you must replace y with (-y) and x with (-x) and the resulting function must be equal to the original function.

\(\displaystyle [(-y)=|(-x)|+2]\neq [y=|x|+2]\)

So there is no symmetry about the origin.

For a function to be symmetrical about the y-axis, it must satisfy \(\displaystyle f(x)=f(-x)\)

\(\displaystyle f(x)=\left |-x \right |+2=\left |x \right |+2\) so there is symmetry about the y-axis

For a function to be symmetrical about the x-axis, it must satisfy \(\displaystyle f(-x)=f(x)\)

\(\displaystyle f(-x)=\left |-x \right |+2\neq -f(x)\) so there is not symmetry about the x-axis

For a function to be symmetrical about the origin, you must replace y with (-y) and x with (-x) and the resulting function must be equal to the original function.

\(\displaystyle \left [ (-y)=\left |-x \right |+2 \right ]\neq \left [y=\left |x \right |+2 \right ]\)

So there is no symmetry about the origin, and the credited answer is "symmetry about the y-axis".

For a function to be symmetrical about the y-axis, it must satisfy \(\displaystyle f(x)=f(-x)\)

\(\displaystyle f(-x)=\sqrt{4+x}=-f(x)\) so there is not symmetry about the y-axis

For a function to be symmetrical about the x-axis, it must satisfy \(\displaystyle f(-x)=f(x)\)

\(\displaystyle f(-x)=\sqrt{4+x}=-f(x)\) so there is symmetry about the x-axis

For a function to be symmetrical about the origin, you must replace y with (-y) and x with (-x), and the resulting function must be equal to the original function.

\(\displaystyle \left ( -x \right )=-(-y)^{2}+4\neq -y^{2}+4\)

So there is no symmetry about the origin, and the answer is Symmetrical about the x-axis.

For a function to be even, it must satisfy the equality \(\displaystyle f(x)=f(-x)\)

Likewise, if a function is even, it is symmetrical about the y-axis

\(\displaystyle f(-x)=(-x)^{3}+5(-x)-7=-x^{3}-5x-7\neq f(x)\)

Therefore, the function is not even, and so the answer is No.

If the roots of the equation are at x= -4 and x=3, then we can work backwards to see what equation those roots were derived from. If we factored a quadratic equation and obtained the given solutions, it would mean the factored form looked something like:

\(\displaystyle (x+4)(x-3)=0\)

Because this is the form that would yield the solutions x= -4 and x=3. If we work backwards and multiply the factors back together, we get the following quadratic equation:

\(\displaystyle x^2+x-12=0\)

We can make a quadratic polynomial with \(\displaystyle x_1, x_2\) by mutiplying the linear polynomials they are roots of, and multiplying them out.

Start

\(\displaystyle (x-(1+i))(x-(1-i))\)  

Distribute the negative sign

\(\displaystyle (x-1-i)(x-1+i)\) 

FOIL the two polynomials. This means multiply the firsts, then the outers, followed by the inners and lastly, the last terms.

\(\displaystyle x^2-x+ix-x+1-i-ix+i-i^2\)    

\(\displaystyle x^2-x+ix-x+1-i-ix+i+1\)    

\(\displaystyle i^2 = -1\)

Simplify

\(\displaystyle x^2-2x+2\)    

Since we know the solutions of the equation, we know that:

\(\displaystyle (x-4.5)(x-2.3)=0\)

We simply carry out the multiplication on the left side of the equation to get the quadratic equation.

\(\displaystyle (x-4.5)(x-2.3)=x^2-4.5x-2.3x+10.35=x^2-6.8x+10.35\)

These two points tell us that the quadratic function has zeros at \(\displaystyle x=0\), and at \(\displaystyle x=6\).

These correspond to the linear expressions \(\displaystyle x\), and \(\displaystyle (x-6)\).

Expand their product and you arrive at the correct answer.

If the quadratic is opening up the coefficient infront of the squared term will be positive. Thus we get:

\(\displaystyle f(x)=x(x-6)=x^2-6x\).

If the quadratic is opening down it would pass through the same two points but have the equation:

\(\displaystyle f(x)=-x(x-6)=-x^2+6x\).

Since only \(\displaystyle f(x)=x^{2}-6x\) is seen in the answer choices, it is the correct answer.