Quadratic Functions – Exercises Solutions
For each of the following quadratic functions, find the x-intercepts, the f(x)-intercept
(y-intercept), and the vertex. Sketch a graph showing and labeling each of these items.
1. f(x) = x2 – 6x + 9
The function f is a quadratic function. Therefore its graph is a parabola. The parabola
opens up because the leading coefficient 1 > 0. The y-intercept is obtained by setting
x = 0 and solving for f(x). So the y-intercept is the constant term 9.
The x-intercepts are the real zeros of the function f. Zeros of a function
are always found by solving the equation resulting from f(x) = 0. In this
example the zeros are the solutions of 0 = x2 – 6x + 9. This quadratic
equation in one variable may easily be solved with factoring and The
Zero Factor Property as demonstrated below:
x2 – 6x + 9 = 0
(x – 3)(x – 3) = 0 which should viewed as (x – 3)2 = 0
x=3
Therefore 3 is the only zero of the function f. This zero of f is a real
number. Therefore it is an x-intercept. The only way a parabola can
have only one x-intercept is for that x-intercept to be the vertex of the
parabola.
The graph of f is shown in Fig. 1.
2. f(x) = x2 + 7x + 4
The function f is a quadratic function. Therefore its graph is a parabola. The parabola
opens up because the leading coefficient 1 > 0. The y-intercept is obtained by setting
x = 0 and solving for f(x). So the y-intercept is the constant term 4.
The x-intercepts are the real zeros of the function f. Zeros of a function are always
found by solving the equation resulting from f(x) = 0. In this example the zeros are the
solutions of 0 = x2 + 7x + 4. This quadratic equation in one variable is not easily solved
with factoring and The Zero Factor Property but the solution may be obtained from the
b  b2  4ac
quadratic formula x 
as demonstrated below:
2a
x
7  72  (4)(1)(4) 7  33

2(1)
2
Therefore
7  33
2
and
7  33
are the two zeros of the function f. Both of these
2
zeros of f are real numbers. Therefore they both are x-intercepts.
 b
b 
 7
 7  
The vertex of f is  , f      , f    .
 2 
 2a  2a    2
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The range element associated with
7
is
2
2
49 49
 7   7 
 7 
f
   2   7 2   4  4  2  4
2

 




49 2   49   4   4  49 16 33





4
4
4
4
4
2  2 
 7
33 
Therefore the vertex is  ,  
4 
 2
The graph of f is shown in Fig. 2.
3. f(x) = 8x2 – 2x – 7
The function f is a quadratic function. Therefore its graph is a parabola. The parabola
opens up because the leading coefficient 8 > 0. The y-intercept is obtained by setting
x = 0 and solving for f(x). So the y-intercept is the constant term – 7.
The x-intercepts are the real zeros of the function f. Zeros of a function are always
found by solving the equation resulting from f(x) = 0. In this example the zeros are the
solutions of 0 = x2 – 6x + 9. This quadratic equation in one variable is not easily solved
with factoring and The Zero Factor Property but the solution may be obtained from the
b  b2  4ac
quadratic formula x 
as demonstrated below:
2a
x
2  (2)2  (4)(8)(7) 2  228 2  2 57 1  57



2(8)
16
16
8
Therefore
1  57
8
and
1  57
are the two zeros of the function f. Both of these zeros
8
of f are real numbers. Therefore they both are x-intercepts.
1
 1 
The vertex of f is  , f    .
 8  8 
The range element associated with
1
is
8
2
1 2 56 57
1
1
1
f    8   2   7   

8
8
8
8 8 8
8
 
 
 
1
57 
Therefore the vertex is  ,  
8 
8
The graph of f is shown in Fig. 3.
Notice that the graph seems to show one of the x-intercepts as (1,0).
That erroneous conclusion stems from the fact that the resolution of
the graph does not show the difference between 1 and 1.0687.
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4. f(x) = 5x2 – 14x – 3
The function f is a quadratic function. Therefore its graph is a parabola. The parabola
opens up because the leading coefficient 5 > 0. The y-intercept is obtained by setting
x = 0 and solving for f(x). So the y-intercept is the constant term – 3.
The x-intercepts are the real zeros of the function f. Zeros of a function are always
found by solving the equation resulting from f(x) = 0. In this example the zeros are the
solutions of 0 = 5x2 – 14x – 3. This quadratic equation in one variable may easily be
solved with factoring and The Zero Factor Property as demonstrated below:
5x2 – 14x – 3 = 0
(x – 3)(5x + 1) = 0
By the Zero Factor Property x = 3 or x 
1
5
1
are the two zeros of the function f. Both of these
5
zeros of f are real numbers. Therefore they both are x-intercepts.
Note that the factoring gives an alternate form for the rule of f.
Therefore 3 and

1


f(x) = 5x2 – 14x – 3 = (x  3)  x   . This can be used to simplify
5
computations when finding range elements.
 14
 14    7  7  
    , f    .
 10  10    5  5  
7
is
The range element associated with
5
  7  15 
64
7 7
  7 
 8 
f      3   5    1  
7  1  
8 




5
5 5
  5 
 5 
  5 
The vertex of f is 
,f 
7
64 
Therefore the vertex is  ,  
5 
5
The graph of f is shown in Fig. 4.
5. f(x) = x2 + 10 x + 25
Begin by observing the right side of the rule is a perfect square so that
f(x) = x2 + 10 x + 25 = (x + 5)2
The function f is a quadratic function. Therefore its graph is a parabola. The parabola
opens up because the leading coefficient 1 > 0. The y-intercept is obtained by setting
x = 0 and solving for f(x). So the y-intercept is the constant term 25.
The x-intercepts are the real zeros of the function f. Zeros of a function are
always found by solving the equation resulting from f(x) = 0. In this example
the zeros are the solutions of 0 = (x + 5)2. Clearly – 5 is the only solution for
this equation and it is a real number. Therefore it is an x-intercept. The only
way a parabola can have only one x-intercept is for that x-intercept to be the
vertex of the parabola.
The graph of f is shown in Fig. 5.
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10. f(x) = – x2 – 2x + 3
Begin by observing the right side of the rule is factorable so that
f(x) = – x2 – 2x + 3 = (-1)(x2 + 2x – 3) = (-1)(x + 3)(x – 1)
The function f is a quadratic function. Therefore its graph is a parabola. The parabola
opens down because the leading coefficient – 1 < 0. The y-intercept is obtained by
setting x = 0 and solving for f(x). So the y-intercept is the constant term 3.
The x-intercepts are the real zeros of the function f. Zeros of a function are always
found by solving the equation resulting from f(x) = 0. In this example the zeros are the
solutions of 0 = (x + 3)(x – 1). This quadratic equation in one variable may easily be
solved with factoring and The Zero Factor Property as demonstrated below:
(x + 3)(x – 1) = 0
By the Zero Factor Property x = – 3 or x = 1
Therefore – 3 and 1 are the two zeros of the function f.
Both of these zeros of f are real numbers. Therefore
they both are x-intercepts.
 b
b 
The vertex of f is  , f      1, f(1) .
 2a  2a  
The range element associated with -1 is
f(-1) = (–1)( – 1+ 3)( – 1 – 1) = (– 1)(2)( – 2) = 4
Therefore the vertex is (– 1, 4)
The graph of f is shown in Fig. 10.
Fig. 10A shows both the graph of f and its opposite.
Observe the x-intercepts are the same. Why?
Observe every point on the graph of f is reflected through the x-axis to obtain the graph
of its opposite. Why?
Observe the graph of the opposite of f opens in the opposite direction as does the graph
of f. Why?
11. f(x) = x2 + 1
The function f is a quadratic function with positive leading coefficient. Therefore its
graph is a parabola which opens up. The y-intercept is obtained by setting x = 0 and
solving for f(x). So the y-intercept is the constant term 1.
The x-intercepts are the real zeros of the function f. Zeros of a function are always
found by solving the equation resulting from f(x) = 0. In this example the zeros
are the solutions of 0 = x2 +1. The only solutions for this equation are the
complex numbers  i . Therefore the function f has no x-intercepts.
Comparison of f with the squaring function reveals the graph of f to be the
graph of the squaring function shifted (translated) up by one unit. Therefore
the vertex of f is (0,1)
The graph of f is shown in Fig. 11.
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15. f(x) = – 5x2 + 5x
Begin by observing the right side of the rule is factorable so that
f(x) = – 5x2 + 5x = – 5x(x – 1)
The function f is a quadratic function. Therefore its graph is a parabola. The parabola
opens down because the leading coefficient – 5 < 0. The y-intercept is obtained by
setting x = 0 and solving for f(x). So the y-intercept is 0.
The x-intercepts are the real zeros of the function f. Zeros of a function are always
found by solving the equation resulting from f(x) = 0. In this example the zeros are the
solutions of 0 = – 5x2 + 5x. This quadratic equation in one variable may easily be
solved with The Zero Factor Property as demonstrated below:
– 5x(x – 1) = 0
By the Zero Factor Property x = 0 or x = 1
Therefore 0 and 1 are the two zeros of the function f. Both of these
zeros of f are real numbers. Therefore they both are x-intercepts.
 b
b 
1
1 
The vertex of f is  , f      , f    .
 2a  2a    2  2  
1
The range element associated with
is
2
5
1
11
  5 1
f    5     1         
4
2
22
  2 2
1 5
Therefore the vertex is  , 
2 4
The graph of f is shown in Fig. 15.
18. f(x) = 3x2 + 12x + 16
The function f is a quadratic function. Therefore its graph is a parabola. The parabola
opens up because the leading coefficient 3 > 0. The y-intercept is obtained by setting
x = 0 and solving for f(x). So the y-intercept is the constant term 16.
The x-intercepts are the real zeros of the function f. Zeros of a function are
always found by solving the equation resulting from f(x) = 0. In this example
the zeros are the solutions of 0 = 3x2 + 12x + 16. The discriminant of this
quadratic is
b2  4ac  122  (4)(3)(16)  (12)(12)  (12)(16)  (12)(12  16)  12(4)  0 .
Therefore the quadratic has two complex solution and no real solutions. The
 b
b 
func tion has no x-intercepts. The vertex of f is  , f      2, f  2   . The
 2a  2a  
range element associated with -2 is f(– 2) = 3(– 22) + 12(– 2) + 16 = 4.
Therefore the vertex is  2, 4 
The graph of f is shown in Fig.18.
20) f(x) = x2 + 8x + 7
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Begin by observing the right side of the rule is factorable so that
f(x) = x2 + 8x + 7 = (x + 7)(x + 1)
The function f is a quadratic function. Therefore its graph is a parabola. The parabola
opens up because the leading coefficient 1 > 0. The y-intercept is obtained by setting
x = 0 and solving for f(x). So the y-intercept is the constant term 7.
The x-intercepts are the real zeros of the function f. Zeros of a function are always
found by solving the equation resulting from f(x) = 0. In this example the zeros are the
solutions of 0 = x2 + 8x + 7. This quadratic equation in one variable may easily be
solved with The Zero Factor Property as demonstrated below:
(x + 7)(x + 1) = 0
By the Zero Factor Property x = – 7 or x = – 1
Therefore – 7 and – 1 are the two zeros of the function f. Both of these zeros of f are
real numbers. Therefore they both are x-intercepts.
The line of symmetry is the vertical line x = – 4 midway between the x-intercepts.
The vertex of f is on the line of symmetry so it is  4, f(4) .
The range element associated with – 4 is
f(– 4) = (– 4 + 7)( – 4 + 1) = (3)( – 3)= – 9.
Therefore the vertex is (– 4, – 9)
The graph of f is shown in Fig. 20
.
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