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The Humongous Book of Algebra Problems by W. Michael Kelley

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Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
488
Parabolas
Vertex, axis of symmetry, focus, and directrix
Note: Problems 22.1–22.3 refer to the equation y = x
2
– 2x + 5.
22.1 Write the equation of the parabola in standard form.
The standard form of a parabola containing an x
2
-term is y = a(xh)
2
+ k, where
(h,k) is the vertex and a is a real number. Subtract 5 from both sides of the
equation in order to isolate the x-terms right of the equal sign.
y – 5 = x
2
– 2x
To complete the square on the right side of the equation, add 1 to both sides of
the equation.
Factor the quadratic expression.
y – 4 = (x – 1)
2
Solve for y.
y = (x – 1)
2
+ 4
Note: Problems 22.1–22.3 refer to the equation y = x
2
– 2x + 5.
22.2 Identify the vertex and axis of symmetry.
The standard form of a parabola containing an x
2
-term is y = a(xh)
2
+ k.
According to Problem 22.1, the standard form of the parabola is y = (x – 1)
2
+ 4.
Therefore, a = 1, h = 1, and k = 4.
The vertex of the parabola is (h,k) = (1,4). The axis of symmetry, the vertical
line that passes through the vertex, has equation x = h; the axis of
symmetry for this parabola is x = 1.
You’ll need to
complete the
square for the
expression x
2
– 2x.
Move all of the terms
that dont contain x
to the left side of
the equation.
To complete
the square, divide
the x-coefcient by
two (–2 ÷ 2 = –1) and
square the result
([1]
2
= 1). To keep
everything balanced, you
have to add 1 to the
left and right sides of
the equation.
This is the
number in front of
the parentheses. When
there is no number,
a = 1.
This is the
OPPOSITE of the
number in parentheses.
The opposite of –1 is 1.
This is the
number added
to the squared
quantity.
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
489
Note: Problems 22.1–22.3 refer to the equation y = x
2
– 2x + 5.
22.3 Graph the parabola.
To transform the equation y = x
2
into the standard form equation
y = (x – 1)
2
+ 4, subtract 1 from the input x (which shifts the graph of y = x
2
one
unit to the right) and add 4 (which shifts the graph up four units). The graph
is presented in Figure 22-1.
Figure 22-1: The graph of y = x
2
– 2x + 5 has vertex (1,4) and axis of symmetry x = 1.
Note: Problems 22.4–22.6 refer to the equation x = y
2
– 6y + 7.
22.4 Write the equation of the parabola in standard form.
The standard form of a parabola containing a y
2
-term is x = a(yk)
2
+ h, where
(h,k) is the vertex and a is a real number. Subtract 7 from both sides of the
equation to isolate the y-terms right of the equal sign.
x – 7 = y
2
– 6y
Complete the square on the right side of the equation by adding 9 to both sides.
Factor the quadratic expression.
x + 2 = (y – 3)
2
Solve for x.
x = (y – 3)
2
– 2
If youre not
sure how to graph
a parabola using
transformations, look
at Problems 16.31 and
16.32.
Divide the
y-coefcient by
two (6 ÷ 2 = –3)
and square the result
([–3]
2
= 9). Add that
number to both sides
of the equation to
keep everything
balanced.

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