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Parabolas and Functions: Vertex Formula and One-to-One Functions, Assignments of Algebra

Weber State University Algebra

The concepts of finding the vertex of parabolas using the formula and identifying one-to-one functions. It includes examples of calculating the vertex of various parabolas and determining if functions are one-to-one. The document also includes the inverse of some functions and their domains.

Typology: Assignments

Pre 2010

Uploaded on 07/23/2009

koofers-user-nvp 🇺🇸

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Section 9.5

9. The vertex of y= (x+ 3)2−4 is at (−3,4).

21. y=−2x2

33. f(x) = −1

2(x+ 1)2+ 2

35. f(x) = 2(x−2)2−3

Section 9.6

5. f(x) = x2+8x+10 The vertex formula gives that the vertex is at x=−b

2a=−4 and y=f(−4) =

−6. This thing opens upward (like the letter U).

25. f(x) = −2x2+ 4x−5. The vertex formula gives that the vertex is at x=−b

2a= 1 and

y=f(1) = −3. This basically means that you can rewrite f(x) as f(x) = −2(x−1)2−3.

29. x=−1

5y2+ 16y+ 11 The vertex formula gives that the vertex is at y=−b

2a=16

2/5= 40 and

x=−1

5402+16(40)+11 = 331. This basically means that you can rewrite it as x=−1

5(y−40)2+331.

This is a parabola that opens to the left.

39. s(t) = −16t2+ 64t+ 3. Using the vertex formula tells us that the vertex is at the point (2,67).

This basically means that the function is s(t) = −16(t−2)2+ 67. Now since (t−2)2is always

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Section 9.

The vertex of y = (x + 3)^2 − 4 is at (− 3 , 4).
y = − 2 x^2
f (x) = −^12 (x + 1)^2 + 2
f (x) = 2(x − 2)^2 − 3

Section 9.

f (x) = x^2 +8x+10 The vertex formula gives that the vertex is at x = − 2 ab = −4 and y = f (−4) = −6. This thing opens upward (like the letter U).
f (x) = − 2 x^2 + 4x − 5. The vertex formula gives that the vertex is at x = − 2 ab = 1 and y = f (1) = −3. This basically means that you can rewrite f (x) as f (x) = −2(x − 1)^2 − 3.
x = −^15 y^2 + 16y + 11 The vertex formula gives that the vertex is at y = − 2 ab = (^216) / 5 = 40 and x = −^15 402 +16(40)+11 = 331. This basically means that you can rewrite it as x = −^15 (y−40)^2 +331. This is a parabola that opens to the left.
s(t) = − 16 t^2 + 64t + 3. Using the vertex formula tells us that the vertex is at the point (2, 67). This basically means that the function is s(t) = −16(t − 2)^2 + 67. Now since (t − 2)^2 is always

nonnegative, it follows that −16(t − 2)^2 is always less than or equal to 0. So no matter what input you choose, there is no way the output can be bigger than 67. Therefore, the max height is when t = 2 and s(t) = 67 feet.

Section 10.

Only the graph in A is one-to-one since it is the only one that passes the horizontal line test.
This function is one-to-one, and its inverse is {(6, 3), (10, 2), (12, 5)}.
If g(x) = √x − 3, we think of this as y = √x − 3. The inverse can be found by writing x = √y − 3 and solving for y. This gives y = x^2 + 3. But the range of g(x) is [0, ∞), so the domain of g−^1 (x) is [0, ∞).
(a) f (3) = 2^3 = 8 and (b) f −^1 (8) = 3 because of the fact that f (3) = 8.
(a) It is a bit hard to tell for sure, but it appears that this function is one-to-one. (b) The graph is shown to the right. This is the result of reflecting the graph (from the book) over the line y = x.

Section 10.

y = 4−x^ is the same as y =

)x .

Rewrite 16^2 x+1^ = 64x+3^ as 16^2 x+1^ = (16^3 /^2 )x+3^ = 16(3/2)x+(9/2), and thus we must solve 2 x + 1 = (3/2)x + (9/2), and this gives (1/2)x = 7/2. Therefore, x = 7.
To solve (3/2)x^ = 8/27, just note that 2^3 = 8 and 3^3 = 27. So the answer is x = −1.

Parabolas and Functions: Vertex Formula and One-to-One Functions, Assignments of Algebra

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