Quadratic Functions and Parabolas
A function that can be put in the form f (x) = ax 2 + bx + c , where a ≠ 0 , is called a quadratic function.
This arrangement of a quadratic function is called the general form. The number a is called the leading
coefficient.
Examples: Identify the constants a, b, and c in each of the following quadratic functions.
€
€
(a) f (x) = 6x 2 −13x + 5
(b) f (x) = −3x 2 + 48x
(c) f (x) = 2x 2 −18
a = 6, b = −13, and c = 5
a = −3, b = 48, and c = 0
a = 2, b = 0, and c = −18
€
€
€
€
€
€
€
€
€
€
The graph of the quadratic function
f (x) = ax 2 + bx + c , where a€≠ 0 is a U-shaped graph called a
€
parabola. The parabola opens€up ∪ if a > 0 and opens down ∩€if a < 0 . The tip of the parabola is
called its vertex. If the parabola opens up, the vertex is the lowest point on the parabola and the ycoordinate of the vertex is the minimum value of the function. If the parabola opens down, the vertex
€
is the highest point on the €
parabola and the y-coordinate of the vertex is the maximum value of the
€ €
€ €b
function. The x-coordinate of the vertex is given by x = − . The y-coordinate of the vertex is
2a
⎛ b ⎞
given by f ⎜ − ⎟ . The graph of a quadratic function is symmetric about the vertical line that passes
⎝ 2a ⎠
through its vertex. (This means that if the graph
€ is folded on the vertical line through the vertex, the two
halves of the parabola will lie exactly on top of each other.) This vertical line is called the axis of
b
symmetry and its equation is x = − , the same as the x-coordinate of the vertex.
€
2a
2
Example: For f (x) = −x + 4 x + 5 , give (a) the x-intercept(s), (b) the y-intercept, (c) both coordinates of
the vertex, and (d) the equation of the axis of symmetry. (e) Graph f (x) .
Solution:
€
(a) Substituting 0 for f (x) , we have
€
0 = −x 2 + 4 x + 5 , which we solve by factoring.
€
0 = −1(x 2 − 4 x − 5)
0 = −1(x − 5)(x
€ +1)
x − 5 = 0 or x +1 = 0
x = 5 or x = −1.
So the x-intercepts are (5,0) and (-1,0).
(b) Substituting 0 for x, we get
f (0) = −0 2 + 4(0) + 5 = 5 ,
so the y-intercept is (0,5).
(c) The x-coordinate of the vertex is given by
b = − 4 = − 4 = 2.
x = − 2a
2(−1)
−2
To get the y-coordinate of vertex, we
substitute this value of x in the function’s
formula and get
f (2) = −(2) 2 + 4(2) + 5 = −4 + 8 + 5 = 9 .
So the vertex is (2,9).
(d) The equation of the axis of symmetry is given by the same formula as the x-coordinate of the vertex,
so the axis of symmetry is x = 2 .
(e) The graph of f (x) = −x 2 + 4 x + 5 is the open down parabola shown above.
€ 2015)
(Thomason – Spring
€
p. 1 of 2
The equation for a quadratic function can also be put in the form f (x) = a(x − h) 2 + k , called the vertex
form. As in the general form, the graph is a parabola that opens up if a > 0 and opens down if a < 0 .
The vertex of the parabola is (h,k) and the axis of symmetry is the vertical line x = h .
Example: For f (x) = 2(x − 3) 2 − 8 , give (a) the x-intercept(s), (b) the y-intercept, (c) both coordinates of
€
the vertex, and (d) the equation of the axis of symmetry. (e)€Graph f (x) .
€
Solution: (a) Substituting 0 for f (x) , we get
€
2
0 = 2(x
€ − 3) − 8
8 = 2(x − 3) 2
€
€
€
€
€
€
4 = (x − 3) 2
±2 = x − 3
3±2 = x
so x = 3 − 2 = 1 and x = 3 + 2 = 5 . In other
words, the x-intercepts are (1,0) and (5,0).
(b) Substituting 0 for x, we get
f (x) = 2(0 − 3) 2 − 8 = 2(9) − 8 = 18 − 8 = 10
so the y-intercept is (0,10).
(c) Comparing f (x) = 2(x − 3) 2 − 8 to the
vertex form f (x) = a(x − h) 2 + k , we see that
h = 3 and k = −8 . So the vertex is (3,−8) .
(d) The
€ axis of symmetry is the vertical line
through the vertex, so its equation is x = 3.
€
(e) The graph of f (x) = €
2(x − 3) 2 − 8 is the open up parabola shown above.
€
To convert a quadratic function
from vertex form to general form, square the binomial and simplify.
2
Example: Convert f (x) = −(x − 5) + 25 to standard form.
€
Solution: f (x) = −(x 2 −10x + 25) + 25 = −x 2 +10x − 25 + 25 = −x 2 +10x , which is the general form.
To convert
€ a quadratic function from general form to vertex form, use completing the square. The
notation looks simpler if we first replace f (x) with y. Then subtract the constant term, if any, from both
€
sides. Next, if the x 2 coefficient is not already one, divide both sides by that coefficient. This is needed
because completing the square only “works” when the squared term’s coefficient is 1. Now complete the
square on the right and remember that any operation done to one side of the equation must also be done
€
to the other side. Next, factor the perfect square trinomial that you’ve created on the right side into the
square of€a binomial. Then solve the equation for y and finally replace y with f (x) . These steps are
illustrated in the following example.
Example: Convert f (x) = 3x 2 −18x + 7 to vertex form.
Solution: y = 3x 2 −18x + 7
y − 7 = 3x 2 −18x
1 y − 7 =€x 2 − 6x
3
3
€
1 y − 7 + 9 = x 2 − 6x + 9
3
3
1 y − 7 + 27 = x 2 − 6x + 9
3
3
3
1 y + 20 = (x − 3) 2
3
3
1 y = (x − 3) 2 − 20
3
3
(continued in next column)
(continued in next column)
€
y = 3(x − 3) 2 − 20
f (x) = 3(x − 3) 2 − 20
€
€
€
(Thomason – Spring 2015)
p. 2 of 2