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(x – h)

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(x – h)

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 don’t contain x

to the left side of

the equation.

To complete

the square, divide

the x-coefcient 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(y – k)

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 you’re not

sure how to graph

a parabola using

transformations, look

at Problems 16.31 and

16.32.

Divide the

y-coefcient 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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