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:

Starting with $\displaystyle y = x^{2}$

$\displaystyle (x-2)^{2}$ moves the parabola $\displaystyle x^{2}$ by $\displaystyle 2$ units to the right.

Similarly $\displaystyle (x+3)^{2}$ moves the parabola by $\displaystyle 3$ units to the left.

Hence the correct answer is option $\displaystyle 3$.

When we look at the function we see that the highest power of the function is a 3 which means it is an "odd degree" function. This means that the right and left side of the function will approach opposite directions. *Remember O for Odd and O for opposite. 

In this case we also have a negative sign associated with the highest power portion of the function - this means that the function is flipped. 

Both of these combine to make this an "odd negative" function. 

Odd negative functions always have the right side of the function approaching down and the left side approaching up. 

We represent this mathematically by saying that as x approaches negative infinity (left side), the function will approach positive infinity: 

$\displaystyle As\ x\rightarrow\ -\infty; f(x)\rightarrow\infty;$

...and as x approaches positive infinity (right side) the function will approach negative infinity:

$\displaystyle As\ x\rightarrow\infty; f(x)\rightarrow-\infty;$

$\displaystyle To\ find\ the\ x-intercepts\ plug\ in\ zero\ for\ f(x)$

$\displaystyle f(x)=(x-1)(x+2)(x-5)^2$

$\displaystyle 0=(x-1)(x+2)(x-5)^2$

Then set each factor equal to zero, if any of the ( ) equal zero, then the whole thing will equal zero because of the zero product rule. 

$\displaystyle x-1=0\ \ \ \ \ \ \ \ \ x+2=0\ \ \ \ \ \ \ \ \ \ \ (x-5)^2=0$

$\displaystyle x=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=5$

As a polynomial function, the graph of $\displaystyle P(x)$ is continuous. By the Intermediate Value Theorem, if $\displaystyle P(a) < M < P(b)$ or $\displaystyle P(b) < M < P(a)$, then there must exist a value $\displaystyle c \in (a,b)$ such that $\displaystyle P(c)= M$. 

Set $\displaystyle a = 0$ and $\displaystyle b= 1$. It is not true that $\displaystyle P(0) < 0< P(1)$, so the Intermediate Value Theorem does not prove that there exists $\displaystyle c \in (0,1 )$ such that $\displaystyle P(c)= 0$. However, it does not disprove that such a value exists either. For example, observe the graphs below:

Both are polynomial graphs fitting the given conditions, but the only the equation graphed at right has a zero on $\displaystyle c \in (0,1)$.

The graph of a quadratic function $\displaystyle f(x)$ has an $\displaystyle x$-intercept at any point $\displaystyle (x,0 )$ at which $\displaystyle f(x) = 0$, so we set the quadratic expression equal to 0:

$\displaystyle x^{2} + 7x - 9 = 0$

Since the question simply asks for the number of $\displaystyle x$-intercepts, it suffices to find the discriminant of the equation and to use it to determine this number. The discriminant of the quadratic equation 

$\displaystyle ax^{2} + bx+ c = 0$

is 

$\displaystyle b^{2} - 4ac$.

Set $\displaystyle a = 1, b= 7, c = -9$, and evaluate:

$\displaystyle b^{2} - 4ac = 7^{2} - 4 (1)(-9)= 49 - (-36) =49 + 36 = 85$

The discriminant is positive, so the $\displaystyle f(x)$ has two real zeroes - and its graph has two $\displaystyle x$-intercepts.

The graph of the quadratic function $\displaystyle f(x) = ax^{2} + bx+ c$ is a parabola with its vertex at the point with coordinates

$\displaystyle \left ( -\frac{b}{2a}, f \left (-\frac{b}{2a} \right ) \right )$.

Set $\displaystyle a = -4, b = 12$; the $\displaystyle x$-coordinate is 

$\displaystyle -\frac{b}{2a} = - \frac{12}{2 \cdot (-4)} = - \frac{12}{-8} = \frac{3}{2} = 1\frac{1}{2}$

Evaluate $\displaystyle f \left ( \frac{3}{2} \right )$ by substitution:

$\displaystyle f \left ( \frac{3}{2} \right )= -4 \left ( \frac{3}{2} \right )^{2}+ 12 \left ( \frac{3}{2} \right ) - 9$

$\displaystyle = -4 \left ( \frac{9}{4} \right ) + 12 \left ( \frac{3}{2} \right ) - 9$

$\displaystyle = -9 +18 - 9$

$\displaystyle = 0$

The vertex has 0 as its $\displaystyle y$-coordinate; it is therefore on an axis.

As a polynomial function, the graph of $\displaystyle P(x)$ is continuous. By the Intermediate Value Theorem (IVT), if $\displaystyle P(a) < M < P(b)$ or $\displaystyle P(b) < M < P(a)$, then there must exist a value $\displaystyle c \in (a,b)$ such that $\displaystyle P(c)= M$.

Setting $\displaystyle M = 0, a = 0, b=1$,  and examining the first condition, the above becomes:

if $\displaystyle P(0) < 0 < P(1)$, then there must exist a value $\displaystyle c \in (0, 1)$ such that $\displaystyle P(c)= 0$ - or, restated, $\displaystyle P(x)$ must have a zero on the interval $\displaystyle (0,1 )$. Since $\displaystyle P(0) = -5$, $\displaystyle P (1) = 17$. the condition holds, and by the IVT, it follows that $\displaystyle P(x)$ has a zero on $\displaystyle (0,1 )$.

As a polynomial function, the graph of $\displaystyle P(x)$ is continuous. By the Intermediate Value Theorem, if $\displaystyle P(a) < M < P(b)$ or $\displaystyle P(b) < M < P(a)$, then there must exist a value $\displaystyle c \in (a,b)$ such that $\displaystyle P(c)= M$.

Setting $\displaystyle a= 0$ and $\displaystyle M = 0$, assuming for now that $\displaystyle b > 0$, and looking only at the second condition, this statement becomes: If $\displaystyle P(b) < 0 < P(a)$, then there must exist a value $\displaystyle c \in (0, b)$ such that $\displaystyle P(c)= 0$ - or, equivalently, $\displaystyle P(x)$ must have a zero on $\displaystyle (0, b)$.

However, the conclusion of this statement is false: $\displaystyle P(x)$ has no zeroes at all. Therefore, $\displaystyle P(b) < 0$ is false, and $\displaystyle P(x)$ has no negative values for any $\displaystyle x >0$. By similar reasoning, $\displaystyle P(x)$ has no negative values for any $\displaystyle x< 0$. Therefore, by the IVT, by way of its contrapositive, we have proved that $\displaystyle P(x)$ is positive everywhere.

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 

$\displaystyle 3=C(4-3)(4+2)$  

$\displaystyle 6C=3$

$\displaystyle C=\frac{1}{2}$