Solving Word Problems Involving Quadratic Functions
Solving Word Problems Involving Quadratic Function
Some of the most important functions in applications are the quadratic functions. The quadratic functions are one of the simplest type of polynomial function, aside from the linear functions. Applications include satellite dishes, projectile paths, economics, geometry, ecology, weather,biology and many others.
A function f is a quadratic function if f(x) = ax2 + bx + c where a, b, and c are real numbers and not equal to 0.. The graph of a quadratic function is a parabola. The most basic aid in graphing a parabola is knowing whether a > 0 (the graph opens upward) or a < 0 (the graph opens downward). The two simplest quadratic functions are f(x) = x2 and g(x) = - x2 .
In this hub, I feature two interesting problems with their complete solution.
Problem Number One :
Jack Pott dives off the high diving board. His distance from the surface of the water
varies quadratically with the number of seconds that have passed since he left the
(a) His distance at time 1, 2 and 3 seconds since he left the board are 24, 18, 2
meters above the water respectively. Write a particular quadratic equation
expressing the distance in terms of time.
(b) How high is the diving board ?
(c ) When does he hit the water ?
(a) In finding the quadratic equation three points can be derived from the given :
(1, 24 ) ; (2, 18) ; (3, 2);
by letting the time in seconds as x-coordinates and the distance above the
water as y-coordinates.
The quadratic equation has a standard form ax2 + bx + c = 0.
Using the given points above, we must solve for a, b, and c.
From the point (1, 24) we can derive an equation by substituting 1 to x’s and
24 to y, The equation formed is : a + b + c = 24 let this be equation #1.
From point (2, 18) we can derive this equation : 4a + 2b + c = 18 let this be
From point ( 3, 2 ) we can derive this equation : 9a + 3b + c = 2 let this be
We will now solve for a, b, and c using elimination method. by subtraction.
Using equation 1 and equation 2 to eliminate c ;
a + b + c = 24
-4a - 2b - c = -18
-3a - b = 6 let this be equation 4
Using equation 2 and equation 3 to eliminate c
4a + 2b + c = 18
-9a - 3b + c = - 2
-5a - b = 16 let this be equation 5
Solving for a and b using equation 4 and equation 5
-3a –b = 6
+5a +b = -16
2a = -10
a = -5
substituting a = -5 in equation 4
-3(-5) – b = 6
15 – b = 6
b = 15 – 6
b = 9
substituting a = -5 b = 9 in equation 1 to solve for c
-5 + 9 + c = 24
c = 24 – 4
c = 20
Therefore the quadratic equation we are looking for is -5x2 + 9x + 20 = 0.
(b) How high is the diving board ?
In order to solve for the height of the diving board we must solve for the vertex of the parabola y = -5x2 + 9x + 20.. This parabola opens downward since a is negative.
Actually, the vertex is the maximum point of the parabola. The height of the diving board is the y-coordinate of the vertex. Solve first for h or x-coordinate of the vertex . The formula to solve for h is -b/2a. Therefore h = -9/(2) (-5) = 9/10.
Substitute 9/10 to every x in the given quadratic function y = -5x2 + 9x + 20.
y = -5 (9/10)2 + 9 (9/10) + 20
y = -81/20 + 81/10 + 20
y = 481/20 = 24.05 meters.
( c ) When does he hit the water ?
y = 0, when the diver hits the water so in order to solve for this we must let
-5x2 + 9x + 20 = 0 and solve for x using quadratic formula.
The quadratic formula is given by :
X = ( -b + - SQRT(b^2-4ac) )/2a
X = (-9 + - SQRT(81 – (4)(-5)(20)) ) /2(-5)
X = (-9 +- SQRT(481) )/-10
X = ( -9 – 21.93 ) /-10
X = -30.3/-10
X = 3.03 seconds
Problem Number Two :
A satellite dish in storage has parabolic cross sections and is resting on its
vertex.. A point on the rim is 4 feet high and is 6 feet horizontally from the
vertex. How high is a point which is 3 feet horizontally from the vertex?
Plotting this parabola on a rectangular coordinate plane with vertex at (0, 0 ), we can derive other two points based from the given . Two points on the rim are
(-6, 4) and (6, 4).
Now let us derive the equation of the parabola using points (0, 0) ; (-6, 4), (6, 4).
C = 0
Using point (-6, 4 ) and (6, 4)
4 = a (-6)^2 + (-6)b + c
4 = 36 a -6b + c
4 = 36 a + 6b + c
8 = 72 a
a = 1/9
Solving for b ;
4 = 36 (1/9) + 6b
4 = 4 + 6b
B = 0
The equation of the parabola is y = 1/9 x^2
Y = 1/9 (3)^2
Y = 1/9(9) = 1
The point which is 3 feet horizontally from the vertex is 1 foot high.
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