The $\displaystyle x$-intercept(s) of the graph of $\displaystyle f(x)$ are the point(s) at which it intersects the $\displaystyle x$-axis. The $\displaystyle y$-coordinate of each is 0,; their $\displaystyle x$-coordinate(s) are those value(s) of $\displaystyle x$ for which $\displaystyle f(x) = 0$, so set up, and solve for $\displaystyle x$, the equation:

$\displaystyle f(x)= 0$

$\displaystyle 2 \cdot 10 ^{x} - 8 = 0$

Add 8 to both sides:

$\displaystyle 2 \cdot 10 ^{x} - 8 + 8 = 0 + 8$

$\displaystyle 2 \cdot 10 ^{x} = 8$

Divide both sides by 2:

$\displaystyle \frac{2 \cdot 10 ^{x}}{2} =\frac{ 8}{2}$

$\displaystyle 10 ^{x} =4$

Take the common logarithm of both sides to eliminate the base:

$\displaystyle \log 10 ^{x} =\log 4$

$\displaystyle x=\log 4$

Let $\displaystyle g(x) = e^{x}$. This function is defined for any real number $\displaystyle x$, so the domain of $\displaystyle g(x)$ is the set of all real numbers. In terms of $\displaystyle g(x)$,

$\displaystyle f(x) = 2g(x-3) - 7$

Since $\displaystyle g(x)$ is defined for all real $\displaystyle x$, so is $\displaystyle g(x-3)$; it follows that $\displaystyle f(x)$ is as well. The correct domain is the set of all real numbers.

Since a positive number raised to any power is equal to a positive number, 

$\displaystyle e^{x-3} > 0$

Applying the properties of inequality, we see that

$\displaystyle 2 \cdot e^{x-3} > 2 \cdot 0$

$\displaystyle 2 e^{x-3} > 0$

$\displaystyle 2 e^{x-3} -7 > 0 - 7$

$\displaystyle 2 e^{x-3} -7 > - 7$

$\displaystyle f(x) > - 7$,

and the range of $\displaystyle f(x)$ is the set $\displaystyle (-7, \infty )$.

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.