Form the quadratic equation of the parabola from the tunnel and graph it together with the measurements of the train, to see if the train can be drawn inside the parabola.
First, turn the tunnel into a graph, using x axis as the ground and y axis as the axis of symmetry
Proceed with forming the equation of the parabola
Start by identifying the x and y intercepts from the graph
p
=
-4
x-intercept
q
=
4
x-intercept
y
=
10
y-intercept
x
=
0
Value of x at the y-intercept
Now, slot these values into the Intercept Form and solve for a
y
=
a(x-p)(x-q)
Intercept Form
10
=
a(0+4)(0-4)
Substitute values
10
=
a(4)(-4)
10
=
-16a
10÷(-16)
=
-16a÷(-16)
-58
=
a
a
=
-58
Substitute a and the x intercepts into the Intercept Form
y
=
a(x-p)(x-q)
Intercept Form
y
=
-58(x-4)(x-4)
Substitute values
y
=
-58(x2-16)
This is the equation for the parabola
This time, turn the train into a graph
Similar to the tunnel, use x axis as the ground and y axis as the axis of symmetry
Substitute x=2.6 to the equation of the parabola and solve for y
If y is greater that 5.9 (height of the train), the train will fit in the tunnel
y
=
-58(x2-16)
Equation of the Parabola
y
=
-58(2.62-16)
Substitute x=2.6
y
=
-58(6.76-16)
y
=
-58(-9.24)
y
=
5.775
5.775 is the height of the tunnel at x=2.6
Since 5.775<5.9, the train will not fit through the tunnel
No
Question 2 of 5
2. Question
Find x given that the area of the rectangle below is 36
Completing the square is done by producing a square of a binomial on the left side of the equal sign. This method is useful when no two rational numbers solve the equation.
Area of a Rectangle
A=breadth×length
Substitute the given values into the Area Formula, then solve for x by Completing the Square
First, slot the given values into the Area Formula to form a quadratic equation
A=36
breadth=x
length=x+4
A
=
breadth×length
Area of a Rectangle
36
=
x(x+4)
Substitute values
x(x+4)
=
36
x2+4x
=
36
Proceed with taking the coefficient of the x term, dividing it by two and then squaring it.
x2+4x
=
36
Coefficient of the x term
4÷2
=
2
Divide it by 2
(2)2
=
4
Square
This number will make the left side a perfect square.
Add 4 to both sides of the equation to keep the balance.
x2+4x
=
36
x2+4x+4
=
36+4
Add 4 to both sides
x2+4x+4
=
40
Now, transform the left side into a square of a binomial by factoring or using cross method.
(x+2)(x+2)
=
40
(x+2)2
=
40
Finally, take the square root of both sides and continue solving for x.
(x+2)2
=
40
√(x+2)2
=
√40
Take the square root
x+2
=
±2√10
Square rooting a number gives a plus and minus solution
x+2-2
=
±2√10-2
Subtract 2 from both sides
x
=
-2±2√10
Simplify
The roots can also be written individually
=
-2+2√10
=
4.325
=
-2-2√10
=
-8.325
Length cannot be a negative value. Therefore, x=4.325
Completing the square is done by producing a square of a binomial on the left side of the equal sign. This method is useful when no two rational numbers solve the equation.
Substitute the given values to the Area Formula to form a quadratic equation, then solve for x by completing the square
First, slot the given values into the Area Formula to form a quadratic equation
A=45
base=x+6
height=x
A
=
12bh
Area of a Triangle
45
=
12(x+6)(x)
Substitute values
45×2
=
12(x+6)(x)×2
Multiply both sides to 2
90
=
(x+6)x
90
=
x2+6x
x2+6x
=
90
Proceed with taking the coefficient of the x term, dividing it by two and then squaring it.
x2+6x
=
90
Coefficient of the x term
6÷2
=
3
Divide it by 2
(3)2
=
9
Square
This number will make the left side a perfect square.
Add 9 to both sides of the equation to keep the balance.
x2+6x
=
90
x2+6x+9
=
90+9
Add 9 to both sides
x2+6x+9
=
99
(x+3)2
=
99
Finally, take the square root of both sides and continue solving for x.
(x+3)2
=
99
√(x+3)2
=
√99
Take the square root
x+3
=
±3√11
Square rooting a number gives a plus and minus solution
x+3-3
=
±3√11-3
Subtract 3 from both sides
x
=
-3±3√11
Simplify
The roots can also be written individually
=
-3+3√11
=
6.9499
=
-3-3√11
=
-12.9499
Length cannot be a negative value. Therefore, x=6.9499
x=6.9499
Question 5 of 5
5. Question
Using only 60m of fencing, find the maximum possible area for a paddock. This paddock is against a barn, which means only 3 sides need to be fenced.
Assign a variable for one side of the paddock and create a quadratic equation using the Area formula. Then, find the maximum area with the help of the axis of symmetry.
First, assign a variable to each side of the rectangle that needs fencing
The two equal sides can be labelled as x. Since the fence will be 60m long in total, the last side would be 60-2x
Form a quadratic equation by substituting values to the Area Formula
breath=x
length=60-2x
A
=
breadth×length
Area Formula
A
=
x(60-2x)
Substitute values
A
=
60x-2x2
A
=
-2x2+60x
Next, solve for the axis of symmetry of the equation
A=-2x2+60x
a=-2b=60c=0
x
=
−b2a
Axis of Symmetry
x
=
−602(−2)
Substitute values
x
=
-60-4
x
=
15
This means that the side x should be 15m long for the paddock to have the maximum area
Finally, substitute x=15 to the given equation to get the maximum area