Precalc Review, any and all help appreciated!

**precalcstudent1994** · December 29th 2010, 12:46 PM

I really need all the help I can get. Please and thank you!!

1. State the domain of f(x) = x+3 / x^2 + x + 2

2. State the domain of f(x) = square root of x^2 - 9

3. Given f(x) = x^3 + x^2 + 1 and g(x) = 2x, find f(g(x))

4. If f(x) = 4x + 3, find f^-1(x)

5. What is the slope and y intercept of the line graphed by the equation y = 3/4x -1

6. Given the points A(-1, 1) and B(4, 13), find the slope of the line AB.

7. Find the length of line segment AB, using the points given in question 6.

8. Write the equation of the line which passes through the points (3, 1) and (-2,4)

9. What is the slope of the line perpendicular to the line 2x + 3y - 6 = 0?

10. Which conic section is graphed by the equation 9x^2 + 16y^2 + 54x - 32y - 47 = 0? Put the equation into standard form and graph the conic section, labeling the center point and points of intersections with the axes.

11. Find the solutions to the equation: y = 3x^2 - 4x + 4

12. Write the function as a product of linear factors: f(x) = x^4 - 2x^3 + x^2 - 8x -12

13. List the discontinuities of the given function and identify each one as either a point discontinuity or a vertical asymptote. f(x) = (x + 2)(x - 3) / (x - 3)(x + 4)

14. Does the given function have a horizontal asymptote? If yes, what is the horizontal asymptote? f(x) = 2x^2 - 3x + 5 / 3x^2

15. Find the partial fraction decomposition of f(x) = 9x^2 - 24x - 57 / (x + 1)(x + 3)(x - 5)

16. Find the value of log 4 (39)

17. Solve for x: log 2 (x + 3) + log 2 (x - 4) = 3

18. Find the sum and identify the series as arithmetic or geometric.

Precalc Review, any and all help appreciated!-latex_sum.gif

(5 on top), n = 1, and (4n + 3) as formula

19. Determine whether the infinite geometric series converges. If it does, find its sum.

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(infinity symbol on top), j =1, and 3(1 / 4)^j

20. Solve the oblique triangle. 31degrees in lover left corner, 22degrees in lower right corner, right leg meaurement of 12, bottom leg = y, left leg = x

21. Find the six trig ratios for the angle whose terminal side contains the point (-2,5). Then find the angle measure.

22. Write the component form of the vector AB given the points A (-2,5) and B (1,3). Then find the magnitude of AB. (AB is a line segment)

23. Find the rectangular coordinates of the polar point (3, 40degrees).

24. Sketch the graph of the polar equation r = 3sin5(theta)

25. Sketch the graph of the parametric equation x = 3cost, y = sin2t

**pickslides** · December 29th 2010, 12:56 PM

Hi there, welcome to the forum.

I suggest you read the forum rules and then re-post some of these questions.

In general it is recommended you only post a few questions per thread with your attempts at a solution. This will give us (the helpers) and you (the learner) the best chance to gain an understanding.

**Soroban** · December 29th 2010, 02:27 PM

Hello, precalcstudent1994!

$\text{10. Identity the conic section: }\:9x^2 + 16y^2 + 54x - 32y - 47 \:=\: 0$
$\text{Put the equation into standard form and graph the conic section,}$
$\text{labeling the center and the vertoces.}$

$\text{We have: }\;9x^2 + 54x \quad + 16y^2 - 32y \quad =\; 47$

$\text{Factor: }\;9(x^2 + 6x \qquad) + 16(y^2 - 2y \qquad) \:=\:47$

$\text{Complete the square: }\;9(x^2 + 6x {\bf+ 9 - 9}) + 16(y^2 - 2y \:{\bf+ 1 - 1}) \:=\:47$

. . . . . $9(x^2+6x+9) - 81 + 16(y^2 - 2y + 1) - 16 \:=\:47$

. . . . . . . . $9(x+3)^2 + 16(y-1)^2 \:=\:144$

$\displaystyle \text{Divide by 144: }\;\frac{(x+3)^2}{16} + \frac{(y-1)^2}{9} \;=\;1$

This is an ellipse: . $a = 4,\:b = 3$

Center: $(-3,1)$

Ends of major axis: . $(1,1),\:(-7,1)$

Ends of minor axis: . $(-3,4),\:(-3,-2)$

$\text{25. Sketch the graph: }\:\begin{Bmatrix}x &=& 3\cos\theta & [1] \\y &=& \sin2\theta & [2] \end{Bmatrix}$

It turned out to be simpler to simply plot some points.

The graph has a "butterfly" shape: . $\begin{array}{c}\bigg\langle\!\bigg\rangle\!\!\big g\langle\!\bigg\rangle \end{array}$ . (rounded)

I eliminated the parameter and got an ugly result.

From [1]: . $\cos\theta \,=\,\dfrac{x}{3}\;\;[3]$

From [2]: . $y \:=\:2\sin\theta\cos\theta$

Substitute [3]: . $y \:=\:2\sin\theta\left(\dfrac{x}{3}\right) \quad\Rightarrow\quad \sin\theta \:=\:\dfrac{3y}{2x}\;\;[4]$

. . $\begin{array}{ccccccc}\text{Square [4]:} & \dfrac{9y^2}{4x^2} &=& \sin^2\!\theta & [5]\\ \\[-3mm] \text{Square [3]:} & \dfrac{x^2}{9} &=& \cos^2\!\theta & [6]\end{array}$

Add [5] and [6]: . $\dfrac{9y^2}{4x^2} + \dfrac{x^2}{9} \;=\;\underbrace{\sin^2\!\theta + \cos^2\!\theta}_{\text{This is 1}}$

We have: . $\displaystyle \frac{9y^2}{4x^2} + \frac{x^2}{9} \:=\:1 \quad\Rightarrow\quad 81y^2 + 4x^4 \:=\:36x^2$

Hence: . $y^2 \:=\:\dfrac{4x^2(9-x^2)}{81} \quad\Rightarrow\quad y \:=\:\pm\dfrac{2x}{9}\sqrt{9-x^2}$

**Jameson** · December 29th 2010, 02:42 PM

Originally Posted by pickslides

Hi there, welcome to the forum.

I suggest you read the forum rules and then re-post some of these questions.

In general it is recommended you only post a few questions per thread with your attempts at a solution. This will give us (the helpers) and you (the learner) the best chance to gain an understanding.

Perfectly said. Thread closed. Next time post just one question and show some work.

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