quadratic functions problems with detailed solutions are presented along with graphical interpretations of the solutions.
Review Vertex and Discriminant of Quadratic Functions
the graph of a quadratic function written in the form
f(x) = a x ^{2} + b x + c
has a vertex at the point (h , k) where h and k are given by
h =  b / (2 a) and k = f(h) = c  b ^{2} / (4 a)
If a > 0, the vertex is a minimum point and the minimum value of the quadratic function f is equal to k. This minimum value occurs at x = h.
If a < 0, the vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This maximum value occurs at x = h.
The quadratic function f(x) = a x ^{2} + b x + c can be written in vertex form as follows:
f(x) = a (x  h) ^{2} + k
The discriminant D of the quadratic equation: a x ^{2} + b x + c = 0 is given by D = b^{2}  4 a c
If D = 0 , the quadratic equation a x ^{2} + b x + c = 0 has one solution and the graph of f(x) = a x ^{2} + b x + c has ONE xintercept.
If D > 0 , the quadratic equation a x ^{2} + b x + c = 0 has two real solutions and the graph of f(x) = a x ^{2} + b x + c TWO two xintercepts.
If D > 0 , the quadratic equation a x ^{2} + b x + c = 0 has two complex solutions and the graph of f(x) = a x ^{2} + b x + c has NO xintercept.
Problems with Solutions
Problem 1
The profit (in thousands of dollars) of a company is given by.
P(x) = 5000 + 1000 x  5 x^{2}
where x is the amount ( in thousands of dollars) the company spends on advertising.
a) Find the amount, x, that the company has to spend to maximize its profit.
b) Find the maximum profit Pmax.
Solution to Problem 1
Problem 2
An object is thrown vertically upward with an initial velocity of V_{o} feet/sec. Its distance S(t), in feet, above ground is given by
S(t) = 16 t ^{2} + v_{o} t.
Find v_{o} so that the highest point the object can reach is 300 feet above ground.
Solution to Problem 2
Problem 3
Find the equation of the quadratic function f whose graph passes through the point (2 , 8) and has x intercepts at (1 , 0) and (2 , 0).
Solution to Problem 3

Since the graph has x intercepts at (1 , 0) and (2 , 0), the function has zeros at x = 1 and x =  2 and may be written as follows.
f(x) = a (x  1)(x + 2)

The graph of f passes through the point (2 , 8), it follows that
f(2) =  8

which leads to
 8 = a (2  1)(2 + 2)

expand the right side of the above equation and group like terms
8 = 4 a

Solve the above equation for a to obtain
a =  2

The equation of f is given by
f(x) =  2 (x  1)(x + 2)

Check answer
f(1) = 0
f(2) = 0
f(2) =  2 (2  1)(2 + 2) = 8
Problem 4
Find values of the parameter m so that the graph of the quadratic function f given by
f(x) = x ^{2} + x + 1
and the graph of the line whose equation is given by
y = m x
have:
a) 2 points of intersection,
b) 1 point of intersection,
c) no points of intersection.
Solution to Problem 4

To find the points of intersection, you need to solve the system of equations
y = x ^{2} + x + 1
y = m x

Substitute m x for y in the first equation to obtain
mx = x ^{2} + x + 1

Write the above quadratic equation in standard form.
x ^{2} + x (1  m) + 1 = 0

Find the discriminant D of the above equation.
D = (1  m) ^{2}  4(1)(1)
D = (1  m) ^{2}  4
a)

For the graph of f and that of the line to have 2 points of intersection, D must be positive, which leads to
(1  m) ^{2}  4 > 0

Solve the above inequality to obtain solution set for m in the intervals
( ∞ , 1) U (3 , + ∞)
b)

For the graph of f and that of the line to have 1 point of intersection, D must be zero, which leads to
(1  m) ^{2}  4 = 0

Solve the above equation to obtain 2 solutions for m.
m = 1
m = 3
c)

For the graph of f and that of the line to have no points of intersection, D must be negative, which leads to
(1  m) ^{2}  4 < 0

Solve the above inequality to obtain solution set for m in the interval
(1 , 3)
The graphs of y = 3 x, y =  x and that of the quadratic function f(x) = x ^{2} + x + 1 are shown in the figure below.
Problem 5
The quadratic function C(x) = a x ^{2} + b x + c represents the cost, in thousands of Dollars, of producing x items. C(x) has a minimum value of 120 thousands for x = 2000 and the fixed cost is equal to 200 thousands. Find the coefficients a,b and c.
Solution to Problem 5

Function C is a quadratic function. Its minimum point, which is given as (2000,120) is the vertex of the graph of C. Hence we can write C(x) in vertex form as follows
C(x) = a (x  2000) ^{2} + 120

The fixed cost is the value of C(x) when x = 0. Hence
C(0) = a (0  2000) ^{2} + 120 = 200

Solve for a
a = 80 / 2000 ^{2} = 0.00002

We expand C(x) and identify the coefficients a, b and c.
C(x) = 0.00002 (x  2000) ^{2} + 120 = 0.00002 x ^{2}  0.08 x + 200
a = 0.00002 , b = 0.08 and c = 200.

The graph of C(x) is shown below and we can verify that the minimum point is at (2000,120) and the fixed cost C(0) = 200.
Problem 6
Find the equation of the tangent line to the the graph of f(x) =  x ^{2} + x  2 at x = 1.
Solution to Problem 6

There are at least two methods to solve the above question.
Method 1

Let the equation of the tangent line be of the form
y = m x + b

and we therefore need to find m and b. The tangent line passes through the point
(1 , f(1)) = (1 , 2)

Hence the equation in m and b
 2 = m (1) + b or m + b = 2

To find the point of tangency of the line and the graph of the quadratic function, we need to solve the system
y = m x + b and y =  x ^{2} + x  2

Substitute y by m x + b in the second equation of the system to obtain
m x + b =  x ^{2} + x  2

Write the above equation in standard form
 x ^{2} + x (1  m)  2  b = 0

For the line to be tangent to the graph of the quadratic function, the discriminant D of the above equation must be equal to zero. Hence
D = b ^{2}  4 a c = (1  m)^{ 2}  4 (1) ( 2  b) = 0

which gives
(1  ( 2  b) )^{ 2} + 4 ( 2  b) = 0

Expand, simplify and write the above equation in standard form
b^{ 2} 2 b + 1 = 0
(b + 1)^{ 2} = 0

Solve for b
b =  1

Find m
m =  2  b = 1

The equation of the tangent line is given by
y =  x  1
Graphical interpretation (or checking) is shown below with the graph of y =  x  1 tangent to the graph of f(x) =  x ^{2} + x  2 at x = 1.
Method 2
The second method is based on the concept of the derivative studied in calculus; see question 4 in Calculus Questions with Answers (5)
Questions with SolutionsQuestion 1
Find the equation of the quadratic function f whose graph has x intercepts at (1 , 0) and (3 , 0) and a y intercept at (0 , 4).
Question 2
Find values of the parameter c so that the graphs of the quadratic function f given by
f(x) = x ^{2} + x + c
and the graph of the line whose equation is given by y = 2 x
have:
a) 2 points of intersection,
b) 1 point of intersection,
c) no points of intersection.
Solutions to the Above Questions
Solution to Question 1

The x intercepts of the graph of f are the zero of f(x). Hence f(x) is of the form
f(x) = a (x + 1)(x  3)

We now need to find coefficient a using the y intercept
f(0) = a(0 + 1)(0  3) =  4

Solve for a
a = 4 / 3

Hence
f(x) = (4 / 3) (x + 1)(x  3)
Solution to Question 2

To find the coordinates of the point of intersections of the graphs of f(x) = x ^{2} + x + c
and y = 2 x, we need to solve the system
y = x ^{2} + x + c and y = 2 x

which by substitution , gives the equation
x^{2} + x + c = 2x

Rewrite the above equation in standard form
x^{2}  x + c = 0

Find the discriminant D
D = 1  4 c

Conclusion
If D is positive or c < 1 / 4 , the two graphs intersect at two points.
If D is equal to 0 or c = 1 / 4 , the two graphs intersect (touch) at 1 point.
If D is negative or c > 1 / 4 , the two graphs have no point of intersection.
More References and Links to Quadratic Functions
