Five–Minute Check (over Chapter 6)

Then/Now

New Vocabulary

Key Concept: Standard Form of Equations for Parabolas

Example 1: Determine Characteristics and Graph

Example 2: Real–World Example: Characteristics of Parabolas

Example 3: Write in Standard Form

Example 4: Write Equations Given Characteristics

Key Concept: Line Tangent to a Parabola

Example 5: Find a Tangent Line at a Point

Over Chapter 6

Solve the system of equations, if possible.2x – y – z = 0x + 2y + z = 33x + 4y + 2z = 8

A. (3, 2, 4)

B. (2, –3, 7)

C. (1, 0, 2)

D. (–1, 1, 3)

Over Chapter 6

Solve the system of equations, if possible.y + 2z = 0–x + y = 12x – y + z = –1

A. (0, 4, –2)

B. (–5, 6, –3)

C. (1, 2, –1)

D. (3, 4, –3)

Over Chapter 6

A test has skill problems worth 4 points each and application problems worth 6 points each. You can do a skill problem in 2 minutes and an application problem in 4 minutes. You have 40 minutes to take the test and may answer no more than 16 problems. Assuming you answer all of the attempted problems correctly, how many of each type of problem must you answer to maximize your score?

A. 0 skill and 10 application

B. 0 skill and 16 application

C. 12 skill and 4 application

D. 16 skill and 0 application

You identified, analyzed, and graphed quadratic functions. (Lesson 1–5)

• Analyze and graph equations of parabolas.

• Write equations of parabolas.

• conic section

• degenerate conic

• locus

• parabola

• focus

• directrix

• axis of symmetry

• vertex

• latus rectum

Determine Characteristics and Graph

For (y – 3)2 = –8(x + 1), identify the vertex, focus, axis of symmetry, and directrix. Then graph the parabola.

The equation is in standard form and the squared term is y, which means that the parabola opens horizontally.

Because 4p = –8 and p = –2, the graph opens left.

The equation is in the form (y – k)2 = 4p(x – h), so h = –1 and k = 3.

Use the values of h, k, and p to determine the characteristics of the parabola.

vertex: (–1, 3) (h, k)

Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the general shape of the curve.

focus: (–3, 3) (h + p, k)

directrix: x = 1 x = h – p

axis of symmetry: y = 3 y = k

Determine Characteristics and Graph

Answer: vertex: (–1, 3); focus: (–3, 3); directrix: x = 1; axis of symmetry: y = 3

Determine Characteristics and Graph

For (x + 1)2 = –4(y – 2), identify the vertex, focus, axis of symmetry, and directrix. Then graph the parabola.

vertex: (–1, 2); focus: (–1, 3); directrix: y = 1; axis of symmetry: x = –1

A. B.

C. D.

vertex: (–1, 2); focus: (–1, 1); directrix: y = 3; axis of symmetry: x = –1

vertex: (–1, 2); focus: (–2, 2); directrix: x = 0; axis of symmetry: y = 2

vertex: (–1, 2); focus: (0, 2); directrix: x = –2; axis of symmetry: y = 2

Characteristics of Parabolas

ASTRONOMY The parabolic mirror for the California Institute of Technology’s Hale telescope at Mount Palomar has a shape modeled by y

2 = 2668x, where x and y are measured in inches. What is the focal length of the mirror?

The equation is provided in standard form, and h and k are both zero. Because 4p = 2668, p is 667. So, the location of the focus is (0 + 667, 0) or (667, 0).

The focal length of the mirror is the distance from the vertex to the focus, or 667 inches.

Answer: 667 inches

ASTRONOMY The cross section of the image of a constellation can be modeled by –12(y – 6) = x

2, where x and y are measured in centimeters. What is the focal length of the cross section?

A. 6 centimeters

B. 12 centimeters

C. 3 centimeters

D.

Write in Standard Form

Write x 2 – 8x – y = –18 in standard form. Identify

the vertex, focus, axis of symmetry, and directrix. Then graph the parabola.

x 2 – 8x – y= –18 Original equation

–y = –x2 + 8x – 18 Subtract x 2 – 8x from

each side.

y = x 2 – 8x + 18 Multiply each side

by –1.

y = x 2 – 8x + 16 + 2 Complete the square.

y = (x – 4)2 + 2 Factor.

y – 2 = (x – 4)2 Subtract 2 from eachside.

Write in Standard Form

(x – 4)2 = y – 2 Standard form of aparabola.

Because the x–term is squared and p = 0.25, the graph opens up. Use the standard form to determine the characteristics of the parabola.

vertex: (4, 2) (h, k)

axis of symmetry: x = 4 x = h

focus: (h, k + p)

directrix: y = y = k – p

Write in Standard Form

Graph the vertex, focus, axis, and directrix of the parabola.

Then make a table of values to graph the curve. The curve should be symmetric about the axis of symmetry.

Write in Standard Form

Answer:

Write y 2 + 16x = 55 – 6y in standard form. Identify the vertex,

focus, axis of symmetry, and directrix. Then graph the parabola.

A. (y + 3)2 = –16(x – 4); vertex: (4, –3); focus: (0, –3); directrix: x = 8; axis of symmetry: y = –3

B. (y + 3)2 = –16(x – 4); vertex: (–3, 4); focus: (–3, 0); directrix: y = 8; axis of symmetry: x = 3

C. (y – 3)2 = –16(x – 4); vertex: (4, 3); focus: (0, 3); directrix: x = 8; axis of symmetry: y = 3

D. (y – 3)2 = –16(x – 4); vertex: (4, 3); focus: (8, 3); directrix: x = 0; axis of symmetry: y = 3

Write Equations Given Characteristics

A. Write an equation for and graph a parabola with focus (2, 1) and vertex (–5, 1).

Because the focus and vertex share the same y–coordinate, the graph is horizontal. The focus is (h + p, k), so the value of p is 2 – (–5) or 7. Because p is positive, the graph opens to the right.

Write the equation for the parabola in standard form using the values of h, p, and k.

4p(x – h) = (y – k)2 Standard form

4(7)[x – (–5)] = (y – 1)2 p = 7, h = –5, and k = 1

28(x + 5) = (y – 1)2 Simplify.

Write Equations Given Characteristics

Graph the vertex and focus. Then make a table of values to graph the parabola.

The standard form of the equation is (y – 1)2 = 28(x + 5).

Answer: (y – 1)2 = 28(x + 5)

xxx-new art (both graph and table)

Write Equations Given Characteristics

B. Write an equation for and graph a parabola with vertex (3, –2) and directrix y = –1.

The directrix is a horizontal line, so the parabola opens vertically. Because the directrix lies above the vertex, the parabola opens down.

Use the equation of the directrix to find p.

y = k – p Equation of directrix

–1 = –2 – p y = –1, k = –2

1 = –p Add 2 to each side.

–1 = p Multiply each side by –1.

Write Equations Given Characteristics

Substitute the values for h, k, and p into the standard form equation for a parabola opening vertically.

4p(y – k) = (x – h)2 Standard form

4(–1)[y – (–2)] = (x – 3)2 p = –1, h = 3, and k = –2

–4(y + 2) = (x – 3)2 Simplify.

The equation for the parabola is (x – 3)2 = –4(y + 2).

Use a table of values to graph the parabola.

Write Equations Given Characteristics

Answer: (x – 3)2 = –4(y + 2)

Write Equations Given Characteristics

C. Write an equation for and graph a parabola that has focus (–1, 7), opens up, and contains (3, 7).

Because the parabola opens up, the vertex is (–1, 7 – p). Use the standard form of the equation of a vertical parabola and the point (3, 7) to find the equation.

(x – h)2 = 4p(y – k) Standard form.

[3 – (–1)]2 = 4p[7 – (7 – p)] x = 3, y = 7, h = –1, k = 7 – p

16 = 4p2 Simplify.

4 = p2 Divide each side by 4.

Write Equations Given Characteristics

Because the parabola opens up, the value of p must be positive. Therefore, p = 2. The vertex is (–1, 5), and the standard form of the equation is (x + 1)2 = 8(y – 5).

Use a table of values to graph the parabola.

Answer: (x + 1)2 = 8(y – 5)

±2 = p Take the square root ofeach side.

Write an equation for and graph a parabola with focus (–2, 5) and directrix x = 4.A. –12(x + 2) = (y – 5)2 B. –12(x – 1) = (y – 5)2

C. 12(x – 1) = (y – 5)2 D. 2(x + 2) = (y – 4.5)2

Write an equation for the line tangent to y = x 2 – 2

at (2, 2).

Find a Tangent Line at a Point

The graph opens vertically. Determine the vertex and focus.

y = x 2 – 2 Original equation

y + 2 = x 2 Write in standard form.

Because 4p = 1, p = 0.25, the vertex is (0, –2) and the focus is (0, –1.75). As shown here, we need to determine d, the distance between the focus and the point of tangency, C.

Find a Tangent Line at a Point

This is one leg of the isosceles triangle.

Find a Tangent Line at a Point

Use d to find A, the endpoint of the other leg of the isosceles triangle.

A = (0, –1.75 – 4.25) or (0, –6)

Slope formula

Find a Tangent Line at a Point

Points A and C both lie on the line tangent to the parabola. Find an equation of this line.

y – y1 = m(x – x1) Point–slope form

y – 2 = 4(x – 2) m = 4, y1 = 2, and x1 = 2

y – 2 = 4x – 8 Distributive Property

y = 4x – 6 Add 2 to each side.

Answer: y = 4x – 6

Write an equation for the line tangent to y 2 = 4x + 4

at (0, 2).

A. y = 2x + 2

B. y = –x + 2

C. y = x + 2

D. y = –2x + 2