name: \underline{\hspace{3cm}} unit 4: solving quadratic equations math ii unit 4 exam #4 date: \underline{\hspace{3cm}} \
q1) a football is kicked from the ground and travels at a rate of 18 meters per second. the function represents the balls height above the ground at x seconds. based on this function model: $f(x) = 18x - 6x^2$ what is footballs highest point and how long did it take to get there? \
q.3) using the factoring method solve for this quadratic equation. $x^2 - 10x + 24 = 0$ \
q.5) solve by completing the square: $x^2 + 8x + 16 = 100$ \
q.7) solve the equation using the factoring method. (mcas q) $6x^2 + 18x = 0$ \
q.9) find the exact function rule of this graph using the formula $y = a(x + m)(x + n)$ where m, n are the roots. \
(graph omitted here) \
q2) solve by factoring: (mcas q) $5x - 1 = x^2 + 2x + 1$ \
q.4) given the quadratic equation in factored form solve for x? (mcas q) $2x(3x - 4) = 0$ \
q.6) use any method to find the solution to the equation $x^2 + 14x + 49 = 36$. \
q.8) given the graph find the simplest function rule. assume $a = 1$. \
(graph omitted here) \
q.10) solve for x: $(x + 5)^2 = 4$

Question

Accepted Answer

Let's solve Q1 first:

### Q1:
# Explanation:
The function for the football's height is $ F(x) = 18x - 6x^2 $. This is a quadratic function in the form $ ax^2 + bx + c $ (here $ a = -6 $, $ b = 18 $, $ c = 0 $). The time to reach the highest point of a parabola (since $ a < 0 $, it opens downward) is at the vertex. The x - coordinate of the vertex of a quadratic function $ y = ax^2+bx + c $ is given by $ x=-\frac{b}{2a} $.

## Step 1: Identify $ a $ and $ b $
For $ F(x)=- 6x^{2}+18x $, we have $ a=-6 $ and $ b = 18 $.

## Step 2: Calculate the x - coordinate of the vertex
Using the formula $ x=-\frac{b}{2a} $, substitute $ a=-6 $ and $ b = 18 $:
$ x=-\frac{18}{2\times(-6)}=-\frac{18}{-12}=\frac{3}{2} = 1.5 $ seconds.

To find the highest point (the maximum height), substitute $ x = 1.5 $ into the function $ F(x) $:

## Step 3: Find the maximum height
$ F(1.5)=18\times(1.5)-6\times(1.5)^{2} $
First, calculate $ 18\times1.5 = 27 $ and $ (1.5)^{2}=2.25 $, then $ 6\times2.25 = 13.5 $
$ F(1.5)=27-13.5 = 13.5 $ meters.

# Answer:
The football's highest point is $ 13.5 $ meters and it takes $ 1.5 $ seconds to get there.

### Q2:
# Explanation:
We start with the equation $ 5x - 1=x^{2}+2x + 1 $. First, we need to rewrite it in standard quadratic form $ ax^{2}+bx + c = 0 $.

## Step 1: Rearrange the equation
Subtract $ 5x $ and add $ 1 $ to both sides:
$ x^{2}+2x + 1-(5x - 1)=0 $
$ x^{2}+2x + 1 - 5x + 1=0 $
$ x^{2}-3x + 2=0 $

## Step 2: Factor the quadratic
We need two numbers that multiply to $ 2 $ (the constant term) and add up to $ - 3 $ (the coefficient of $ x $). The numbers are $ -1 $ and $ -2 $. So, $ x^{2}-3x + 2=(x - 1)(x - 2)=0 $

## Step 3: Solve for $ x $
Set each factor equal to zero:
$ x - 1=0$ or $ x - 2=0 $
$ x = 1$ or $ x=2 $

# Answer:
$ x = 1 $ or $ x = 2 $

### Q3:
# Explanation:
We have the quadratic equation $ x^{2}-10x + 24 = 0 $. We need to factor it. We look for two numbers that multiply to $ 24 $ and add up to $ - 10 $ (since the middle term is $ -10x $ and the constant term is positive, both numbers are negative). The numbers are $ -4 $ and $ -6 $.

## Step 1: Factor the quadratic
$ x^{2}-10x + 24=(x - 4)(x - 6)=0 $

## Step 2: Solve for $ x $
Set each factor equal to zero:
$ x - 4=0$ or $ x - 6=0 $
$ x = 4$ or $ x = 6 $

# Answer:
$ x = 4 $ or $ x = 6 $

### Q5:
# Explanation:
We have the equation $ x^{2}+8x + 16 = 100 $. The left - hand side is a perfect square trinomial. Recall that $ (a + b)^{2}=a^{2}+2ab + b^{2} $, and for $ x^{2}+8x + 16 $, we have $ a=x $, $ 2ab = 8x\Rightarrow b = 4 $, so $ x^{2}+8x + 16=(x + 4)^{2} $

## Step 1: Rewrite the equation as a perfect square
$ (x + 4)^{2}=100 $

## Step 2: Take the square root of both sides
$ x + 4=\pm\sqrt{100}=\pm10 $

## Step 3: Solve for $ x $
Case 1: $ x + 4 = 10 $
$ x=10 - 4=6 $

Case 2: $ x + 4=-10 $
$ x=-10 - 4=-14 $

# Answer:
$ x = 6 $ or $ x=-14 $

### Q7:
# Explanation:
We have the equation $ 6x^{2}+18x = 0 $. We can factor out the greatest common factor (GCF) from the two terms. The GCF of $ 6x^{2} $ and $ 18x $ is $ 6x $.

## Step 1: Factor out the GCF
$ 6x(x + 3)=0 $

## Step 2: Solve for $ x $
Set each factor equal to zero:
- If $ 6x=0 $, then $ x = 0 $
- If $ x + 3=0 $, then $ x=-3 $

# Answer:
$ x = 0 $ or $ x=-3 $

### Q4:
# Explanation:
We have the factored form of the quadratic equation $ 2x(3x - 4)=0 $. According to the zero - product property, if $ ab = 0 $, then either $ a = 0 $ or $ b = 0 $.

## Step 1: Apply the zero - product property
Set $ 2x=0 $ or $ 3x - 4=0 $

## Step 2: Solve for $ x $
- For $ 2x = 0 $, divide both sides by $ 2 $: $ x = 0 $
- For $ 3x-4 = 0 $, add $ 4 $ to both sides: $ 3x=4 $, then divide by $ 3 $: $ x=\frac{4}{3} $

# Answer:
$ x = 0 $ or $ x=\frac{4}{3} $

### Q6:
# Explanation:
We have the equation $ x^{2}+14x + 49 = 36 $. The left - hand side is a perfect square trinomial. Since $ x^{2}+14x + 49=(x + 7)^{2} $ (because $ (x + 7)^{2}=x^{2}+14x + 49 $)

## Step 1: Rewrite the equation as a perfect square
$ (x + 7)^{2}=36 $

## Step 2: Take the square root of both sides
$ x + 7=\pm\sqrt{36}=\pm6 $

## Step 3: Solve for $ x $
Case 1: $ x + 7 = 6 $
$ x=6 - 7=-1 $

Case 2: $ x + 7=-6 $
$ x=-6 - 7=-13 $

# Answer:
$ x=-1 $ or $ x=-13 $

### Q10:
# Explanation:
We have the equation $ (x + 5)^{2}=4 $. We take the square root of both sides.

## Step 1: Take the square root of both sides
$ x + 5=\pm\sqrt{4}=\pm2 $

## Step 2: Solve for $ x $
Case 1: $ x + 5 = 2 $
$ x=2 - 5=-3 $

Case 2: $ x + 5=-2 $
$ x=-2 - 5=-7 $

# Answer:
$ x=-3 $ or $ x=-7 $

For Q8 and Q9, we need to analyze the graph. Let's assume the graph in Q8 has a vertex at $ (0,8) $ and a root at $ x=-4 $ (from the graph). Since $ a = 1 $ and the vertex is at $ (h,k)=(0,8) $, the vertex form of a parabola is $ y=a(x - h)^{2}+k $. So $ y=(x - 0)^{2}+8=x^{2}+8 $? Wait, no, if there is a root at $ x = - 4 $, let's check. Wait, maybe the graph in Q8 is a parabola with vertex at $ (0,8) $ and passing through $ (-4,0) $. Let's use the vertex form $ y=a(x - h)^{2}+k $, $ a = 1 $, $ h = 0 $, $ k = 8 $, so $ y=x^{2}+8 $. But if it passes through $ (-4,0) $, $ 0=(-4)^{2}+8=16 + 8=24
eq0 $. Maybe the root is at $ x=-4 $ and the vertex is at $ (0,8) $, so let's use the factored form. If the roots are $ x=-4 $ and $ x = 4 $ (symmetric about the y - axis), then the factored form is $ y=(x + 4)(x - 4)=x^{2}-16 $, but the vertex of $ y=x^{2}-16 $ is at $ (0,-16) $, not $ (0,8) $. There might be a mistake in my assumption. Alternatively, if the vertex is at $ (0,8) $ and it's a downward - opening parabola (since it has a maximum), $ a=-1 $, then $ y=-x^{2}+8 $. Let's check $ x=-4 $: $ y=-(-4)^{2}+8=-16 + 8=-8
eq0 $. Maybe the graph in Q9: Let's assume the roots are $ x=-4 $ and $ x = 0 $ (from the graph, the blue dot is at $ x=-4 $ and the vertex is at $ (0,8) $). The factored form is $ y=a(x + 4)(x - 0) $, $ a = 1 $, so $ y=x(x + 4)=x^{2}+4x $. The vertex of $ y=x^{2}+4x=(x + 2)^{2}-4 $, which is at $ (-2,-4) $, not $ (0,8) $. Maybe more information is needed, but based on the given, we solved the other questions.

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QUESTION IMAGE

Question

Explanation:

Q1:

Step 1: Identify \( a \) and \( b \)

Step 2: Calculate the x - coordinate of the vertex

Step 3: Find the maximum height

Step 1: Rearrange the equation

Step 2: Factor the quadratic

Step 3: Solve for \( x \)

Step 1: Factor the quadratic

Step 2: Solve for \( x \)

Answer:

Q2: