9. use the most appropriate method to find the number of zeros for each function. explain your though process.
a) $y = 8(x + 6)(x - 9)$ (1 mark)
b) $f(x) = -2(x + 4)^2$ (1 mark)
c) $g(x) = 6x^2 + 11x + 23$ (1 mark)
10. the cost, in thousand of dollars, of making $x$ items, in hundreds, is given by $c(x) = -12x - 75$.
the revenue function is given by $r(x) = -6x^2 + 30$.
determine the number of items that should be sold to maximize profit. (3 marks)

Question

Accepted Answer

### 9a)
# Explanation:
## Step1: Recall zero - product property
The zero - product property states that if $ab = 0$, then either $a = 0$ or $b=0$ (or both). For the function $y = 8(x + 6)(x - 9)$, we set $y = 0$. So we have the equation $8(x + 6)(x - 9)=0$.
## Step2: Analyze the factors
Since $8
eq0$, we can use the zero - product property on $(x + 6)(x - 9)=0$. This gives us two equations: $x+6 = 0$ or $x - 9=0$.
Solving $x + 6=0$ gives $x=-6$, and solving $x - 9 = 0$ gives $x = 9$. So there are two distinct zeros.
# Answer:
The number of zeros of $y = 8(x + 6)(x - 9)$ is $2$.

### 9b)
# Explanation:
## Step1: Recall zero - product property for repeated roots
For the function $f(x)=-2(x + 4)^{2}$, we set $f(x)=0$. So we have the equation $-2(x + 4)^{2}=0$.
## Step2: Analyze the factor
Since $-2
eq0$, we consider $(x + 4)^{2}=0$. By the zero - product property (for a square), if $(x + 4)^{2}=0$, then $x+4=0$ (with multiplicity 2). So the root $x=-4$ is a repeated root, but we count the number of zeros (distinct or repeated, but in terms of the number of solutions to $f(x) = 0$, here we have one distinct zero with multiplicity 2. But when we are asked for the number of zeros (in the sense of the number of values of $x$ that make $f(x)=0$, considering multiplicity or not? In the context of quadratic functions, for $y=a(x - h)^{2}$, the zero is $x = h$ (a repeated root). So the number of zeros (counting multiplicity or as a distinct root) - in the case of a perfect square, we have one zero (even though it has multiplicity 2). Wait, actually, the number of real zeros: for $f(x)=-2(x + 4)^{2}$, when we solve $f(x)=0$, we get $(x + 4)^{2}=0$, so $x=-4$ (a single real zero with multiplicity 2). So the number of real zeros is 1.
# Answer:
The number of zeros of $f(x)=-2(x + 4)^{2}$ is $1$.

### 9c)
# Explanation:
## Step1: Use the discriminant formula
For a quadratic function $ax^{2}+bx + c$ ($a
eq0$), the discriminant $\Delta=b^{2}-4ac$. If $\Delta>0$, there are two distinct real zeros; if $\Delta = 0$, there is one real zero; if $\Delta<0$, there are no real zeros. For the function $g(x)=6x^{2}+11x + 23$, we have $a = 6$, $b = 11$, and $c = 23$.
## Step2: Calculate the discriminant
$\Delta=(11)^{2}-4\times6\times23=121-552=-431$. Since $\Delta=-431<0$, the quadratic function $g(x)$ has no real zeros.
# Answer:
The number of real zeros of $g(x)=6x^{2}+11x + 23$ is $0$.

### 10)
# Explanation:
## Step1: Recall the profit function formula
The profit function $P(x)$ is given by $P(x)=R(x)-C(x)$, where $R(x)$ is the revenue function and $C(x)$ is the cost function.
Given $C(x)=-12x - 75$ and $R(x)=-6x^{2}+30$, then $P(x)=(-6x^{2}+30)-(-12x - 75)$.
## Step2: Simplify the profit function
$P(x)=-6x^{2}+30 + 12x+75=-6x^{2}+12x + 105$.
## Step3: Find the vertex of the quadratic function
For a quadratic function $y = ax^{2}+bx + c$ ($a
eq0$), the $x$ - coordinate of the vertex (which gives the maximum or minimum value) is given by $x=-\frac{b}{2a}$. For $P(x)=-6x^{2}+12x + 105$, $a=-6$ and $b = 12$.
So $x=-\frac{12}{2\times(-6)}=-\frac{12}{-12}=1$.
Since $x$ is in hundreds of items, the number of items (in hundreds) that maximizes profit is $x = 1$ hundred items, so the number of items is $1\times100 = 100$ items. Wait, let's check the calculation of the profit function again. Wait, $C(x)$ is the cost function. Wait, the cost function $C(x)=-12x - 75$ - but cost functions are usually non - negative and increasing (since as you produce more, cost should increase). There might be a typo, but assuming the given functions are correct:
$P(x)=R(x)-C(x)=(-6x^{2}+30)-(-12x - 75)=-6x^{2}+12x + 105$.
The $x$ - coordinate of the vertex of $y = ax^{2}+bx + c$ is $x =-\frac{b}{2a}$. Here $a=-6$, $b = 12$, so $x=-\frac{12}{2\times(-6)} = 1$. Since $x$ is in hundreds of items, the number of items is $1\times100=100$ items. But let's verify the profit function:
$P(1)=-6(1)^{2}+12(1)+105=-6 + 12+105=111$ (in thousands of dollars).
If we take $x = 0$, $P(0)=105$ (in thousands of dollars). If we take $x = 2$, $P(2)=-6(4)+12(2)+105=-24 + 24+105 = 105$ (in thousands of dollars). So the vertex is at $x = 1$, which is the maximum.
# Answer:
The number of items (in hundreds) to be sold to maximize profit is $x = 1$ hundred, so the number of items is $100$ items.

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QUESTION IMAGE

Question

Explanation:

9a)

Step1: Recall zero - product property

Step2: Analyze the factors

Step1: Recall zero - product property for repeated roots

Step2: Analyze the factor

Step1: Use the discriminant formula

Step2: Calculate the discriminant

Answer:

9b)