Word problem worksheet on graphing and solving quadratic inequalities with real-world applications.

Problem Analysis and Solution

The image contains several word problems related to quadratic inequalities. Below, I will solve each problem step by step.

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#### Problem 1: HUTS
Task: Write an inequality that describes the shaded region in the graph of the parabola \( y = -\frac{4}{5}(x-1)^2 + 4 \).

Solution:
1. The given equation is \( y = -\frac{4}{5}(x-1)^2 + 4 \).
2. The shaded region is below the parabola, which means \( y \) must be less than or equal to the value of the parabola at any given \( x \).
3. Therefore, the inequality that describes the shaded region is:
\[
y \leq -\frac{4}{5}(x-1)^2 + 4
\]

Answer:
\[
\boxed{y \leq -\frac{4}{5}(x-1)^2 + 4}
\]

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#### Problem 2: DISCRIMINANTS
Task: Consider the equation \( ax^2 + bx + c = 0 \). Assume that the discriminant is zero and that \( a \) is positive. What are the solutions of the inequality \( ax^2 + bx + c > 0 \)?

Solution:
1. The discriminant of the quadratic equation \( ax^2 + bx + c = 0 \) is given by \( \Delta = b^2 - 4ac \).
2. Since the discriminant is zero (\( \Delta = 0 \)), the quadratic equation has exactly one real root. This root can be found using the quadratic formula:
\[
x = \frac{-b}{2a}
\]
3. The quadratic function \( ax^2 + bx + c \) can be rewritten as:
\[
ax^2 + bx + c = a\left(x - \frac{-b}{2a}\right)^2
\]
Since \( a > 0 \), the parabola opens upwards.
4. The inequality \( ax^2 + bx + c > 0 \) holds for all \( x \) except at the vertex (where the parabola touches the x-axis). Therefore, the solution is:
\[
x \neq \frac{-b}{2a}
\]

Answer:
\[
\boxed{x \neq \frac{-b}{2a}}
\]

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#### Problem 3: TOSSING
Task: Gail tosses a tape measure, and its height \( h \) (in feet) above the ground \( t \) seconds after being tossed is given by \( h = -16t^2 + 32t + 5 \). Gail can catch the object anytime it is above 17 feet. How much time does Gail have to try to catch the tape measure?

Solution:
1. We need to find the time intervals when the height \( h \) is greater than 17 feet. This means solving the inequality:
\[
-16t^2 + 32t + 5 > 17
\]
2. Simplify the inequality:
\[
-16t^2 + 32t + 5 - 17 > 0
\]
\[
-16t^2 + 32t - 12 > 0
\]
3. Divide the entire inequality by \(-4\) (and reverse the inequality sign):
\[
4t^2 - 8t + 3 < 0
\]
4. Solve the quadratic equation \( 4t^2 - 8t + 3 = 0 \) using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 4 \), \( b = -8 \), and \( c = 3 \):
\[
t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(3)}}{2(4)}
\]
\[
t = \frac{8 \pm \sqrt{64 - 48}}{8}
\]
\[
t = \frac{8 \pm \sqrt{16}}{8}
\]
\[
t = \frac{8 \pm 4}{8}
\]
\[
t = \frac{12}{8} = \frac{3}{2} \quad \text{or} \quad t = \frac{4}{8} = \frac{1}{2}
\]
5. The roots are \( t = \frac{1}{2} \) and \( t = \frac{3}{2} \). Since the parabola opens upwards (coefficient of \( t^2 \) is positive), the inequality \( 4t^2 - 8t + 3 < 0 \) holds between the roots:
\[
\frac{1}{2} < t < \frac{3}{2}
\]

Answer:
\[
\boxed{\frac{1}{2} < t < \frac{3}{2}}
\]

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#### Problem 4: KIOSKS
Task: Caleb is designing a kiosk by wrapping a piece of sheet metal with dimensions \( x + 5 \) inches by \( 4x + 8 \) inches into a cylindrical shape. Ignoring cost, Caleb would like a kiosk that has a surface area of at least 4480 square inches. What values of \( x \) satisfy this condition?

Solution:
1. When the sheet metal is wrapped into a cylinder, one dimension becomes the circumference of the base, and the other becomes the height.
2. Let the circumference of the base be \( x + 5 \) and the height be \( 4x + 8 \). The radius \( r \) of the base is:
\[
2\pi r = x + 5 \implies r = \frac{x + 5}{2\pi}
\]
3. The surface area \( A \) of the cylinder (excluding the top and bottom) is given by:
\[
A = 2\pi r h
\]
Substituting \( r = \frac{x + 5}{2\pi} \) and \( h = 4x + 8 \):
\[
A = 2\pi \left(\frac{x + 5}{2\pi}\right)(4x + 8)
\]
\[
A = (x + 5)(4x + 8)
\]
4. We need the surface area to be at least 4480 square inches:
\[
(x + 5)(4x + 8) \geq 4480
\]
5. Expand and simplify the inequality:
\[
4x^2 + 8x + 20x + 40 \geq 4480
\]
\[
4x^2 + 28x + 40 \geq 4480
\]
\[
4x^2 + 28x + 40 - 4480 \geq 0
\]
\[
4x^2 + 28x - 4440 \geq 0
\]
6. Divide the entire inequality by 4:
\[
x^2 + 7x - 1110 \geq 0
\]
7. Solve the quadratic equation \( x^2 + 7x - 1110 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 7 \), and \( c = -1110 \):
\[
x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-1110)}}{2(1)}
\]
\[
x = \frac{-7 \pm \sqrt{49 + 4440}}{2}
\]
\[
x = \frac{-7 \pm \sqrt{4489}}{2}
\]
\[
x = \frac{-7 \pm 67}{2}
\]
\[
x = \frac{60}{2} = 30 \quad \text{or} \quad x = \frac{-74}{2} = -37
\]
8. Since \( x \) represents a physical dimension, it must be positive. The quadratic \( x^2 + 7x - 1110 \geq 0 \) holds for \( x \leq -37 \) or \( x \geq 30 \). However, \( x \leq -37 \) is not physically meaningful.

Answer:
\[
\boxed{x \geq 30}
\]

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#### Problem 5: TUNNELS
Task: An architect wants to use a parabolic arch as the entrance of a tunnel. She sketches the plan on a piece of graph paper. She would like the maximum height of the tunnel to be located at \( (4, 4) \), and she would like the origin to be on the parabola as well. Write an equation for the desired parabola.

Solution:
1. The vertex form of a parabola is:
\[
y = a(x - h)^2 + k
\]
where \( (h, k) \) is the vertex. Here, the vertex is \( (4, 4) \), so:
\[
y = a(x - 4)^2 + 4
\]
2. The origin \( (0, 0) \) lies on the parabola. Substitute \( x = 0 \) and \( y = 0 \) into the equation:
\[
0 = a(0 - 4)^2 + 4
\]
\[
0 = 16a + 4
\]
\[
16a = -4
\]
\[
a = -\frac{1}{4}
\]
3. Substitute \( a = -\frac{1}{4} \) back into the vertex form:
\[
y = -\frac{1}{4}(x - 4)^2 + 4
\]

Answer:
\[
\boxed{y = -\frac{1}{4}(x - 4)^2 + 4}
\]

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#### Problem 6: TUNNELS
Task: Write an inequality that describes the region above the parabola \( y = -\frac{1}{4}(x - 4)^2 + 4 \), part of which will be filled in with concrete. Graph this inequality.

Solution:
1. The region above the parabola corresponds to all points where \( y \) is greater than the value of the parabola at any given \( x \). Therefore, the inequality is:
\[
y > -\frac{1}{4}(x - 4)^2 + 4
\]

Answer:
\[
\boxed{y > -\frac{1}{4}(x - 4)^2 + 4}
\]

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Final Answers:

1. \(\boxed{y \leq -\frac{4}{5}(x-1)^2 + 4}\)
2. \(\boxed{x \neq \frac{-b}{2a}}\)
3. \(\boxed{\frac{1}{2} < t < \frac{3}{2}}\)
4. \(\boxed{x \geq 30}\)
5. \(\boxed{y = -\frac{1}{4}(x - 4)^2 + 4}\)
6. \(\boxed{y > -\frac{1}{4}(x - 4)^2 + 4}\)