Master Baruch College MTH 2003 - Final Test SP17 In 30 Minutes Or LESS

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Baruch College MTH 2003 – Final Test SP17 Overview

MTH 2003 is a precalculus course in Baruch College. MTH 2003 is a preliminary mathematics course students must take prior to starting the calculus as well as in quantitative courses in allied disciplines track. This course is a prerequisite for MTH 2205. Baruch College MTH 2003 Final Test has a calculator section. Therefore, the students are required to master TI 89 or TI 92 graphical calculator for this course.

To view the syllabus for Baruch College MTH 2003 CLICK HERE

To get your hands on Baruch College MTH 2003 Last Minute Cram Course, CLICK HERE

To view sample final exam questions for Baruch College MTH 2003 CLICK HERE.

Problem 1

Step 1.

Let’s take a look at the equation $3x-4y+1=0$ . We need to bring it to the form $y=m\ast x+b$ .

$3x-4y+1=0$

$3x+1=4y$

$\frac{3x+1}{4}=y$

$y=\frac{3}{4}x+\frac{1}{4}$

Step 2.

Slopes of perpendicular lines are negative intercepts of each other (e.g. $\frac{2}{3}$ and $-\frac{3}{2}$ , $\frac{4}{7}$ and $-\frac{7}{4}$ , etc.). Hence, the slope of the perpendicular line is $-\frac{4}{3}$ .

The answer is $A$ .

Problem 2

Step 1.

Let’s first find the expression for $f(x+h)$ :

$f(x+h)=$

$11\left(x+h\right)^2=$

$11\left(x+h\right)\ast\left(x+h\right)=$

$11\left(x^2+2xh+h^2\right)=$

$11x^2+22xh+11h^2$

Step 2.

Let’s now plug f(x+h) in the formula of the difference quotient:

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=$

$\frac{11x^2+22xh+11h^2-11x^2}{h}=$

$\frac{22xh+11h^2}{h}=$

$\frac{h(22x+11h)}{h}=22x+11h$

The answer is $C$ .

Problem 3

Step 1.

For a quadratic equation $ax^2+bx+c$ , the highest or lowest point occurs at its $x_{vertex}=-\frac{b}{2a}$ .

For a given expression $y=x^2-10x+13$ ,

$a=1$

$b=-10$

$c=13$

$x_{vertex}=-\frac{b}{2a}=-\frac{-10}{2\ast1}=-\frac{-10}{2}=-\left(-5\right)=5$

Step 2.

To find the y-coordinate of the vertex, simply plug $x_{vertex}$ into the equation.

$y\left(5\right)=$

$\left(5\right)^2-10\ast\left(5\right)+13=$

$25-50+13=-12$

So the coordinates of the vertex is $(5,-12)$ . The answer is $E$ .

Problem 4

Equation of a circle with radius $R$ and origin $(h,k)$ is $\left(x-h\right)^2+\left(y-k\right)^2=R^2$ .

Step 1.

Rearrange the equation to move $x$ and $y$ terms together.

$x^2+y^2+14x-2y=0$

$x^2+14x+y^2-2y=0$

$x^2+14x+__+y^2-2y+__=0$

Step 2.

Find the coefficients to complete the square for $x$ and $y$ :

$x^2+14x+\left(\frac{14}{2}\right)^2+y^2-2y+\left(\frac{2}{2}\right)^2=\left(\frac{14}{2}\right)^2+\left(\frac{2}{2}\right)^2$

$x^2+14x+49+y^2-2y+1=49+1$

$\left(x+7\right)^2+\left(y-1\right)^2=50$

$R^2=50$ and $R=\sqrt{50}$

The answer is $C$ .

Problem 5

Step 1.

Take a look at the highest exponent of the top and the bottom.

The highest exponent of the numerator $x-7$ is $1$ .

The highest exponent of the denominator $x^2-49$ is $2$ .

Since the exponent of denominator is higher, the horizontal asymptote is $y=0$ . The answer is $A$ .

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Step 1.

The revenue function is

$R\left(x\right)=x\ast demand$

$R\left(x\right)=x\left(100-2x\right)=100x-2x^2$

Step 2.

To find the number of items that will maximize revenue, take the first derivative

$R\prime(x)=100-4x$

Step 3.

Set the denominator to $0$ and solve for $x$

$R^\prime\left(x\right)=100-4x=0$

$x=25$

The answer is $D$ .

Problem 7

Step 1.

Find the first derivative of $f\left(x\right)=-4x^2+32x-10$

$f^\prime\left(x\right)=-8x+32$

Step 2.

Set the first derivative to $0$ and solve for $x$ .

$f^\prime\left(x\right)=-8x+32=0$

$x=4$

Step 3.

Plug $x=4$ into the original function to find the y-coordinate

$f\left(x\right)=-4\ast\left(4\right)^2+32\ast4-10=-64+128-10=54$

The point is $(4,54)$ . The answer is $A$ .

Problem 8

Step 1.

Find the first derivative of $f\left(x\right)=\frac{x+1}{x}$ using quotient rule

$f=x+1$ and $f\prime=1$

$g=x$ and $g\prime=1$

Plug values into the formula:

$f^\prime\left(x\right)=\frac{f^\prime\ast g-f\ast g^\prime}{g^2}=$

$\frac{1\ast x-\left(x+1\right)\ast1}{x^2}=$

$\frac{x-(x+1)}{x^2}=\frac{-1}{x^2}$

Step 2.

Plug in $x=2$ to find $f\prime(2)$

$f^\prime\left(x\right)=\frac{-1}{x^2}$

$f^\prime\left(2\right)=\frac{-1}{2^2}=\frac{-1}{4}$

The answer is $D$ .

Problem 9

To find the equation of a line tangent, follow the 4-step process:

Step 1.

Find $f^\prime\left(x\right) of y=x^3-x\$

$f^\prime\left(x\right)=3x^2-1$

Step 2.

Find $f^\prime\left(1\right)\$

$f^\prime\left(1\right)=3\ast\left(1\right)^2-1=3-1=2$

Step 3.

Find $f\left(1\right)\$

$f(1)=\left(1\right)^3-1=1-1=0$

Step 4.

Find the equation of the tangent line by plugging into the following formula:

$y=f^\prime\left(1\right)\ast\left(x-1\right)+f(1)\$

$y=2\left(x-1\right)+0$

$y=2x-2$

Problem 10

In this problem we have to look at the multiple choice answers first. There are $3$ points mentioned:

For $x=0$ , the slope of the function would be negative. Therefore, $f^\prime\left(0\right)<0$ and option $A$ is incorrect.

For $x=-1$ , the slope of the function would be negative. Therefore, $f^\prime\left(-1\right)<0$ and option $B$ is correct, while option $C$ is wrong.

It is still a good idea to double check that other options do not apply.

For $x=1$ , the slope of the function would also be negative. Therefore, $f^\prime\left(1\right)<0$ and option $D$ is incorrect.

The right answer is $B$ .

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Master Baruch College MTH 2003 – Final Test SP17 In 30 Minutes Or LESS

Baruch College MTH 2003 – Final Test SP17 Overview

Problem 1

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Problem 2

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Problem 3

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Problem 4

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Problem 6

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Problem 7

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Problem 8

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Problem 9

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Problem 10

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Baruch College MTH 2003 – Final Test SP17 Overview

Problem 1

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Problem 2

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Problem 3

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Problem 4

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Problem 5

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Problem 6

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Problem 7

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Problem 8

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Problem 9

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Problem 10

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