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

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

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

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 

$\displaystyle 3=C(4-3)(4+2)$  

$\displaystyle 6C=3$

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

The $\displaystyle y$-intercept of the graph of a function $\displaystyle f(x)$ is the point at which it crosses the $\displaystyle y$-axis; its $\displaystyle x$-coordinate is 0, so its $\displaystyle y$-coordinate is $\displaystyle f(0)$. 

$\displaystyle f(x) = -3x^{2}+ 7x + 1$,

so, by setting $\displaystyle x= 0$,

$\displaystyle f(0) = -3 \cdot 0^{2}+ 7\cdot 0 + 1 = 0 + 0 + 1 = 1$,

making $\displaystyle (0, 1)$ the $\displaystyle y$-intercept.

The parabola of an equation of the form $\displaystyle y= ax^{2}+ bx + c$ is vertical, and faces upward or downward depending entirely on the sign of $\displaystyle a$, the coefficient of $\displaystyle x^{2}$. This coefficient, $\displaystyle -4$, is negative; the parabola is concave downward.

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 $\displaystyle M = 0$, $\displaystyle a = 0$ and $\displaystyle b= 1$, this becomes:  If $\displaystyle P(0) < 0 < P(1)$ or $\displaystyle P(1) < 0 < P(0)$, then there must exist a value $\displaystyle c \in (0,1)$ such that $\displaystyle P(c)= 0$ - that is, $\displaystyle P(x)$ must have a zero on $\displaystyle (0,1)$.

However, the question is asking us to use the converse of this statement, which is not true in general. If $\displaystyle P(x)$ has a zero on $\displaystyle (0,1)$, it does not necessarily follow that $\displaystyle P(0) < 0 < P(1)$ or $\displaystyle P(1) < 0 < P(0)$ - specifically, with $\displaystyle P(0) = 4 > 0$, it does not necessarily follow that $\displaystyle P(1) < 0$. A counterexample is the function shown below, which fits the conditions of the problem but does not have a negative value for $\displaystyle P(1)$:

The answer is false.