Rewrite the system as 

\(\displaystyle 2 \cdot 2 ^{x} + 3^{y} = 59\)

\(\displaystyle 2 ^{x } - 3^{y} = -11\)

and substitute \(\displaystyle u\) and \(\displaystyle v\) for \(\displaystyle 2 ^{x }\) and \(\displaystyle 3^{y}\), respectively, to form the system

\(\displaystyle 2u + v= 59\)

\(\displaystyle u- v= -11\)

Add both sides:

\(\displaystyle 2u + v= 5\)

\(\displaystyle \underline{u- v= 43}\)

\(\displaystyle 3u\)        \(\displaystyle =48\)

\(\displaystyle u = 16\).

Now backsolve:

\(\displaystyle 16- v= -11\)

\(\displaystyle - v= - 27\)

\(\displaystyle v= 27\)

Now substitute back:

\(\displaystyle u = 16\)

\(\displaystyle 2 ^{x } = 16\)

\(\displaystyle x = 4\)

and

\(\displaystyle v= 27\)

\(\displaystyle 3^{y}= 27\)

\(\displaystyle y=3\)

\(\displaystyle x+y =4+3 = 7\)

If the graph of \(\displaystyle f (x) = Ax^{2} + 4x- 2\) is concave upward, then \(\displaystyle A > 0\). 

If the graph does not intersect the \(\displaystyle x\)-axis, then \(\displaystyle Ax^{2} + 4x- 2 = 0\) has no real solution, and the discriminant \(\displaystyle 4^{2} - 4 \cdot A \cdot \left ( -2\right )\) is negative:

\(\displaystyle 4^{2} - 4 \cdot A \cdot \left ( -2\right ) < 0\)

\(\displaystyle 16 +8 A < 0\)

\(\displaystyle 16 +8 A - 16 < 0 - 16\)

\(\displaystyle 8 A < - 16\)

\(\displaystyle 8 A \div 8< - 16 \div 8\)

\(\displaystyle A < -2\)

For the parabola to have both characteristics, it must be true that \(\displaystyle A < -2\) and \(\displaystyle A > 0\), but these two events are mutually exclusive. Therefore, the parabola cannot exist.

The graph of \(\displaystyle f(x)= Ax^{2}+Bx+ C\) has as its line of symmetry the vertical line of the equation

\(\displaystyle x=- \frac{B}{2A}\)

Since \(\displaystyle A=7\) in each choice, we want to find \(\displaystyle B\) such that 

\(\displaystyle 7=- \frac{B}{2A} = - \frac{B}{2(7)} = - \frac{B}{14}\)

\(\displaystyle - \frac{B}{14} = 7\)

\(\displaystyle B = 7\cdot (-14)=-98\)

so the correct choice is \(\displaystyle f(x)=7x^{2}-98x+140\).

A horizontal parabola has as its equation, in standard form,

\(\displaystyle x= Ay^{2}+ By+ C\),

with \(\displaystyle A,B,C\) real, \(\displaystyle A\) nonzero.

Its orientation depends on the sign of \(\displaystyle A\). In the equation of a concave-right parabola, \(\displaystyle A\) is positive, so the correct choice is \(\displaystyle x=7y^{2}- 4y+13\).

The graph of a function of the form \(\displaystyle f(x)= Ax^{2}+Bx+C\) - a quadratic function - is a vertical parabola with line of symmetry \(\displaystyle x= - \frac{B}{2A}\). 

The graph of the function \(\displaystyle f(x) = 2x^{2}+ 4x+7\) therefore has line of symmetry 

\(\displaystyle x= - \frac{4}{2 (2)}\), or \(\displaystyle x = -1\)

We examine all four definitions of \(\displaystyle g\) to find one with this line of symmetry.

\(\displaystyle g(x) = 2x^{2}-4x-7\):

\(\displaystyle x= - \frac{-4}{2 (2)}\), or \(\displaystyle x = 1\)

\(\displaystyle g(x) = x^{2}+ 4x+7\):

\(\displaystyle x= - \frac{ 4}{2 (1)}\), or \(\displaystyle x = -2\)

\(\displaystyle g(x) = 2x^{2}+ 8x+7\)

\(\displaystyle x= - \frac{ 8}{2 (2)}\), or \(\displaystyle x = -2\)

\(\displaystyle g(x) = 4x^{2}+8x+7\)

\(\displaystyle x= - \frac{ 8}{2 (4)}\), or \(\displaystyle x = -1\)

Since the graph of the function \(\displaystyle g(x) = 4x^{2}+8x+7\) has the same line of symmetry as that of the function \(\displaystyle f(x) = 2x^{2}+ 4x+7\), that is the correct choice.

We can set the two quadratic expressions equal to each other and solve for \(\displaystyle x\).

\(\displaystyle y= x^{2}+ 2x- 7\) and \(\displaystyle y = 2x^{2}- 6x+5\), so

\(\displaystyle 2x^{2}- 6x+5 = x^{2}+ 2x- 7\)

\(\displaystyle 2x^{2}- 6x+5 -( x^{2}+ 2x- 7)= x^{2}+ 2x- 7 -( x^{2}+ 2x- 7)\)

\(\displaystyle 2x^{2}- 6x+5 - x^{2}- 2x+ 7=0\)

\(\displaystyle x^{2}- 8x+12=0\)

\(\displaystyle (x-2)(x-6) = 0\)

The \(\displaystyle x\)-coordinates of the points of intersection are 2 and 6. To find the \(\displaystyle y\)-coordinates, substitute in either equation:

\(\displaystyle y= x^{2}+ 2x- 7\)

\(\displaystyle y= 2^{2}+ 2 \cdot 2 - 7\)

\(\displaystyle y= 4+4 - 7\)

\(\displaystyle y=1\)

One point of intersection is \(\displaystyle (2,1)\).

\(\displaystyle y= x^{2}+ 2x- 7\)

\(\displaystyle y= 6^{2}+ 2 \cdot 6 - 7\)

\(\displaystyle y= 36+ 12 - 7\)

\(\displaystyle y = 41\)

The other point of intersection is \(\displaystyle (6,41)\).

1 is not among the choices, but 41 is, so this is the correct response.

The \(\displaystyle x\)-intercepts, if any exist, can be found by setting \(\displaystyle f(x) = 0\):

\(\displaystyle f(x)= \frac{1}{7} (x-7)^{2} = 0\)

\(\displaystyle \frac{1}{7} (x-7)^{2} \cdot 7 = 0\cdot 7\)

\(\displaystyle (x-7)^{2} = 0\)

\(\displaystyle x-7 = 0\)

\(\displaystyle x= 7\)

The only \(\displaystyle x\)-intercept is \(\displaystyle (7,0)\).

The \(\displaystyle y\)-intercept can be found by substituting 0 for \(\displaystyle x\):

\(\displaystyle f(0)= \frac{1}{7} (0-7)^{2} = \frac{1}{7} ( -7)^{2}= \frac{1}{7} \cdot 49 = 7\)

The \(\displaystyle y\)-intercept is \(\displaystyle (0,7)\).

The correct set of intercepts is \(\displaystyle \left \{ (7,0), (0,7) \right \}\).

The system of equations can be rewritten as

\(\displaystyle y= x^{2}- 4x+ 7\)

\(\displaystyle y = 4x\).

We can set the two expressions in \(\displaystyle x\) equal to each other and solve:

\(\displaystyle x^{2}- 4x+ 7 = 4x\)

\(\displaystyle x^{2}- 8x+ 7 =0\)

\(\displaystyle (x-1)(x-7)= 0\)

\(\displaystyle x= 1,x=7\)

We can substitute back into the equation \(\displaystyle y = 4x\), and see that either \(\displaystyle y = 4\) or \(\displaystyle y = 28\). The latter value is the correct choice.

I) The orientation of the parabola is determined solely by the sign of \(\displaystyle A\). Since \(\displaystyle A= -5 < 0\), the parabola can be determined to be concave downward.

II and V) The \(\displaystyle x\)-coordinate of the vertex is \(\displaystyle -\frac{B}{2A}\); since you are not given \(\displaystyle B\), you cannot find this. Also, since the line of symmetry has equation \(\displaystyle x=-\frac{B}{2A}\), for the same reason, you cannot find this either.

III) The \(\displaystyle y\)-intercept is the point at which \(\displaystyle x = 0\); by substitution, it can be found to be at \(\displaystyle (0,C)\). \(\displaystyle C\) known to be equal to 9, so the \(\displaystyle y\)-intercept can be determined to be \(\displaystyle (0,9)\).

IV) The \(\displaystyle x\)-intercept(s), if any, are the point(s) at which \(\displaystyle f(x)= Ax^{2}+ Bx+ C = 0\). This is solvable using the quadratic formula

\(\displaystyle x= \frac{-B \pm \sqrt{B^{2}-4AC}}{2A}\)

Since all three of \(\displaystyle A,B,\) and \(\displaystyle C\) must be known for this to be evaluated, and only \(\displaystyle A\) is known, the \(\displaystyle x\)-intercept(s) cannot be identified.

The correct response is I and III only.

The graph of \(\displaystyle f(x)= Ax^{2}+Bx+ C\) has exactly one \(\displaystyle x\)-intercept if and only if 

\(\displaystyle Ax^{2}+Bx+ C = 0\) 

has exactly one solution - or equivalently, if and only if

\(\displaystyle B^{2}-4AC = 0\)

Since in all three equations, \(\displaystyle A = 3,B = -12\), we find the value of \(\displaystyle C\) that makes this statement true by substituting and solving:

\(\displaystyle (12)^{2}-4(3)C = 0\)

\(\displaystyle 144-12C = 0\)

\(\displaystyle 12C = 144\)

\(\displaystyle C= 12\)

The correct choice is \(\displaystyle f(x)= 3x^{2}- 12x + 12\).