3.1 Parabola primer
Solutions
1. Make a rough sketch of the following parabolas, marking the coordinates of the vertex. In each case
there's very little calculation to be done––you can do it by inspection.
(a) y = 3(x–1)(x–7)
(b) y = –2(x+4)(x–1)
(c) y = 5x(x–3)
(d) y = 1–(x+1)(x–3)
(e) y = x(x–4) + 3
(f) y = x2 – 2
(g) y =2(x–2)2
(h) y = (x+2)2 – 2
(i) y = 3–(x–3)2
(j) y = 4–x2
2. Complete the square of each of the following polynomials, make a rough sketch of the graph and mark
on this sketch the coordinates of the vertex. Find the zeros if any.
(a) y = x2 – 4x – 5
(b) y = x2 + 6x + 1
(c) y = 3x2 + 12x – 6
(d) y = –x2 + 6x + 1
(e) y = 4x2 – 1
(f) y = x2 + 5x – 5
(g) y = x(x–8)
3. Either by factoring or by rearranging or by use of the quadratic formula find the roots of the following
quadratic polynomials, or state that none exist. When the roots are irrational, find them precisely in as simple a form as possible, and find also a 2-place decimal approximation.
(a) y = x2 – 4x – 1
(b) y = 2(x–2)2 – 5
(c) y = 2(x–2)2 + 5
(d) y = –x2 + 6x – 1
(e) y = 4x2 – 6x + 5
(f) y = x2 – 2x – 3
(g) y = x(x–8) + 16
(h) y = 2x2 + 5x + 2
(i) y = x2 + 6x + 4
4. In each case find the value of x which makes f(x) either a minimum or a maximum and specify whether
we have a max or a min.
(a) f(x) = (x+1)(x–3)
(b) f(x) = 2 + (x–4)2
(c) f(x) = 4 – (x+4)2
(d) f(x) = x2 – 12x + 27
(e) f(x) = x2 –4x + 5
(f) f(x) = 2 – (x–4)4
*(g) f(x) = [2 + (x–4)2]–1
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5. Prove that x 2  8 x  22 can never be less than 6.
6. Show that the maximum value of 24 x  x 2 is 144.
7. Investigation of the effect of a in the equation y = a(x–c)2 + b .
On the same set of axes draw graphs of the three parabolas:
y = (x–1)2 – 4
y = 4(x–1)2 – 4
y = –(x–1)2 – 4
Describe in simple language the relationship between the three curves.
8. In each case sketch the graphs of a number of members of the family of parabolas, enough to allow us to
see the effect of changes in the parameter a. In each case label each curve with the value of the parameter.
(a) y = ax2 – 1.
(b) y = (x–a)2
(c) y = (x–1)2 + a
(d) y = (x–a)2
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9. For different values of the parameter a, we get different parabolas of the family
y = x(x–a)
(a) Sketch a number of these on a single set of axes, and label each curve with the value of the parameter.
(b) It turns out that the set of vertices of the family of parabolas in (a) lies along a parabola. From considerations of symmetry, deduce what the vertex of the parabola must be. Make a rough sketch of this “parabola
of vertices” and use it to improve your sketch of part (a).
(c) Show algebraically that the set of vertices of the parabolas in (a) lies along a parabola. [This is a fine
and discriminating exercise.]
Solution. (a) The sketch appears at the right.
(b) What makes this argument complicated is that
there are many simple steps each following from the
one before. Here are the steps.
1) the parabola y = x(x–a) is the mirror image in the
y-axis of the parabola y = x(x+a).
2) The whole set of parabolas is bilaterally symmetric about the y-axis.
3) The set of vertices of the parabolas is bilaterally
symmetric about the y-axis.
4) The parabola that passes through the set of vertices (call it the “vertex parabola”) must also be bilaterally symmetric about the y-axis.
5) The vertex of the vertex parabola must be on the
y-axis.
6) The vertex of the vertex parabola must be at the
origin.
Is it clear how each step follows from the previous step? For the
most part yes. To establish 1) we have to take x(x–a) and replace x
by –x, and show that that gives us x(x+a). Okay let’s do it:
(–x)(–x–a) = x(x+a)
And we are done.
To establish 6) we observe that the vertex of x2 is at the origin, so the
origin must lie on the vertex parabola, and since this is on the y-axis,
it must be the vertex of this parabola.
(c) The vertex of x(x–a) is half-way between the zeros at x=a/2. Its
height is (a/2)(a/2–a) = –(a/2)2 = –a2/4. So the general vertex is at
(a/2, a2/4). As a varies, what is the curve that passes through these
points? Well let's describe the point. It's the point whose xcoordinate is a/2 and whose y-coordinate is a2/4. We can simply
write that information as:
x = a/2
y = –a2/4.
Now how to we get the equation for y in terms of x from that? Eliminate a. Alternatively, some students might see directly that if you
square the x and change the sign you get y. By any method you get
the equation:
y = –x2.
This is the equation of the vertex curve and it is indeed a parabola.
It is drawn at the right.
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10. You know that the square of any real number is
at least zero. So if x and y are non-negative real
numbers, then ( x 
y ) 2  0 . Show that this im-
x y
 xy . The left side of this expres2
sion is called the arithmetic mean of two numbers x
and y and right side is referred to as their geometric
mean. You have proved that the arithmetic mean is at
least as great is the geometric mean.
plies that
So the general vertex is at (a/2, a^2/4). As a varies, what is the curve that passes through these
points? That's the problem. Well let's describe the point. It's the point whose x-coordinate is a/2
and We find it by writing the parametric equations:
x = a/2
y = a2/4
and eliminating a to get
y = x2.
This is the equation of the vertex curve and it is indeed a parabola. It is drawn at the right.
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