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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 Lesson Menu

Solve the system of equations, if possible Solve the system of equations, if possible. 2x – y – z = 0 x + 2y + z = 3 3x + 4y + 2z = 8 A. (3, 2, 4) B. (2, –3, 7) C. (1, 0, 2) D. (–1, 1, 3) 5–Minute Check 1

Solve the system of equations, if possible Solve the system of equations, if possible. y + 2z = 0 –x + y = 1 2x – y + z = –1 A. (0, 4, –2) B. (–5, 6, –3) C. (1, 2, –1) D. (3, 4, –3) 5–Minute Check 2

A. 0 skill and 10 application B. 0 skill and 16 application 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 5–Minute Check 3

Analyze and graph equations of parabolas. You identified, analyzed, and graphed quadratic functions. (Lesson 1–5) Analyze and graph equations of parabolas. Write equations of parabolas. Then/Now

conic section axis of symmetry vertex latus rectum degenerate conic locus parabola focus directrix Vocabulary

Key Concept 1

Because 4p = –8 and p = –2, the graph opens left. 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. Example 1

axis of symmetry: y = 3 y = k Determine Characteristics and Graph vertex: (–1, 3) (h, k) focus: (–3, 3) (h + p, k) directrix: x = 1 x = h – p axis of symmetry: y = 3 y = 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. Example 1

Determine Characteristics and Graph Answer: vertex: (–1, 3); focus: (–3, 3); directrix: x = 1; axis of symmetry: y = 3 Example 1

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 Example 1

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 Example 2

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. Example 2

x 2 – 8x – y = –18 Original equation 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 each side. Example 3

(x – 4)2 = y – 2 Standard form of a parabola. Write in Standard Form (x – 4)2 = y – 2 Standard form of a parabola. 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) focus: (h, k + p) directrix: y = y = k – p axis of symmetry: x = 4 x = h Example 3

Graph the vertex, focus, axis, and directrix of the parabola. 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. Example 3

Write in Standard Form Answer: Example 3

Write y 2 + 16x = 55 – 6y in standard form 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 Example 3

4p(x – h) = (y – k)2 Standard form 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. Example 4

The standard form of the equation is (y – 1)2 = 28(x + 5). Write Equations Given Characteristics The standard form of the equation is (y – 1)2 = 28(x + 5). Graph the vertex and focus. Then make a table of values to graph the parabola. Answer: (y – 1)2 = 28(x + 5) xxx-new art (both graph and table) Example 4

Use the equation of the directrix to find p. 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. Example 4

4p(y – k) = (x – h)2 Standard form 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. Example 4

Answer: (x – 3)2 = –4(y + 2) Write Equations Given Characteristics Example 4

(x – h)2 = 4p(y – k) Standard form. 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. Example 4

±2 = p Take the square root of each side. Write Equations Given Characteristics ±2 = p Take the square root of each side. 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) Example 4

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 Example 4

Key Concept 3

Write an equation for the line tangent to y = x 2 – 2 at (2, 2). Find a Tangent Line at a Point Write an equation for the line tangent to y = x 2 – 2 at (2, 2). The graph opens vertically. Determine the vertex and focus. y = x 2 – 2 Original equation y + 2 = x 2 Write in standard form. Example 5

Find a Tangent Line at a Point 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. Example 5

This is one leg of the isosceles triangle. Find a Tangent Line at a Point This is one leg of the isosceles triangle. Use d to find A, the endpoint of the other leg of the isosceles triangle. A = (0, –1.75 – 4.25) or (0, –6) Example 5

y – y1 = m(x – x1) Point–slope form 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. Slope formula 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 Example 5

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 Example 5

End of the Lesson