An intersection of two functions is a point they share in common. A diagram can show all the possible solutions:

Notice that:

\(\displaystyle f(x)\) and \(\displaystyle h(x)\) intersect \(\displaystyle 0\) times

\(\displaystyle f(x)\) and \(\displaystyle g(x)\) intersect \(\displaystyle 1\) time

\(\displaystyle g(x)\) and \(\displaystyle h(x)\) intersect \(\displaystyle 2\) times

The diagram shows that \(\displaystyle I\), \(\displaystyle II\), and \(\displaystyle III\) are all possible. 

Start by visualizing the graph associated with the function \(\displaystyle y=x^2\):

Terms within the parentheses associated with the squared x-variable will shift the parabola horizontally, while terms outside of the parentheses will shift the parabola vertically. In the provided equation, 2 is located outside of the parentheses and is subtracted from the terms located within the parentheses; therefore, the parabola in the graph will shift down by 2 units. A simplified graph of \(\displaystyle y=x^2-2\) looks like this:

Remember that there is also a term within the parentheses. Within the parentheses, 1 is subtracted from the x-variable; thus, the parabola in the graph will shift to the right by 1 unit. As a result, the following graph matches the given function \(\displaystyle y=(x-1)^2-2\) :

\(\displaystyle f(x)\) crosses the \(\displaystyle y\) axis when \(\displaystyle x\) equals 0. So, substitute in 0 for \(\displaystyle x\):

\(\displaystyle f(x)=5x^{5}-3x^{2}+7\)

\(\displaystyle f(0)=5(0)^{5}-3(0)^{2}+7\)

\(\displaystyle f(0)=7\)

The zeros of the parabola are at \(\displaystyle 3\) and \(\displaystyle -2\), so when placed into the formula 

\(\displaystyle y=C(x-z_{1})(x-z_{2})\), 

each of their signs is reversed to end up with the correct sign in the answer. The coefficient can be found by plugging in any easily-identifiable, non-zero point to the above formula. For example, we can plug in \(\displaystyle (4,3)\) which gives 

\(\displaystyle 3=C(4-3)(4+2)\)  

\(\displaystyle 6C=3\)

\(\displaystyle C=\frac{1}{2}\)

We have the following answer choices.

1. \(\displaystyle y(x)=-(x + 1)^3 - 1\)
2. \(\displaystyle y(x)=-(x + 1)^2 - 1\)
3. \(\displaystyle y(x)=-x - 1\)
4. \(\displaystyle y(x)=e^{2x-1}+\frac{1}{9}\)

The first equation is a cubic function, which produces a function similar to the graph. The second equation is quadratic and thus, a parabola. The graph does not look like a prabola, so the 2nd equation will be incorrect. The third equation describes a line, but the graph is not linear; the third equation is incorrect. The fourth equation is incorrect because it is an exponential, and the graph is not an exponential. So that leaves the first equation as the best possible choice.

\(\displaystyle f(x)=\frac{9}{10}x^2-7x+2\)

The highest exponent of the variable term is two (\(\displaystyle x^2\)). This tells that this function is quadratic, meaning that it is a parabola.

The graph below will be the answer, as it shows a parabolic curve.

The way to figure out this problem is by understanding behavior of polynomials.

The sign that occurs before the \(\displaystyle x^8\) is positive and therefore it is understood that the function will open upwards. the "8" on the function is an even number which means that the function is going to be u-shaped. The only answer choice that fits both these criteria is:

Define \(\displaystyle g(x)= f(x) - 7 = x^{5}+ 4x^{2} -7\). Then, if \(\displaystyle g(c)= f(c) - 7 = 0\), it follows that \(\displaystyle f(c) = 7\).

By the Intermediate Value Theorem (IVT), if \(\displaystyle g(x)\) is a continuous function, and \(\displaystyle g(a)\) and \(\displaystyle g(b)\) are of unlike sign, then \(\displaystyle g(c) = 0\) for some \(\displaystyle c \in (a, b)\). As a polynomial, \(\displaystyle g(x)\) is a continuous function, so the IVT applies here.

Evaluate \(\displaystyle g(x)\) for each of the following values: \(\displaystyle \left \{ 1.0, 1.1, 1.2, 1.3, 1.4, 1.5\right \}\)

\(\displaystyle g(1.0) = 1.0^{5}+ 4 \cdot 1.0^{2} -7 = -2\)

\(\displaystyle g(1.1) = 1.1^{5}+ 4 \cdot 1.1^{2} -7 \approx -0.55\)

\(\displaystyle g(1.2) = 1.2^{5}+ 4 \cdot 1.2^{2} -7 \approx 1.24\)

\(\displaystyle g(1.3) = 1.3^{5}+ 4 \cdot 1.3^{2} -7 \approx 3.47\)

\(\displaystyle g(1.4) = 1.4^{5}+ 4 \cdot 1.4^{2} -7 \approx 6.22\)

\(\displaystyle g(1.5) = 1.5^{5}+ 4 \cdot 1.5^{2} -7 \approx 9.59\)

Only in the case of \(\displaystyle (1.1, 1.2)\) does it hold that \(\displaystyle g (x)\) assumes a different sign at both endpoints - \(\displaystyle g(1.1) < 0 < g(1.2)\). By the IVT, \(\displaystyle g(c) = 0\), and \(\displaystyle f(c) = 7\), for some \(\displaystyle c \in (1.1, 1.2)\).