Algebra practice exam questions

In summary: I'm not sure how far I should go with this.)In summary, we discussed formulas for the point-slope formula, difference of squares factorization, discriminant of a quadratic, quadratic formula, consequence of the remainder theorem, distance formula, and midpoint formula. We also solved various problems involving these formulas, including finding the equation of a line, factoring quadratics, finding the value of a variable, and finding the coordinates of points. Finally, we used a graph to discuss finding a possible equation for a given graph.
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MarkFL
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A student I help has been given a set of questions to study for an upcoming exam, and has asked me to work the problems so that they can see the solutions worked out. I decided to post them here so that they may benefit others studying the same material.

Here are some formulas to which I will be referring:

(1) The point-slope formula

A line passing through the point $\left(x_1,y_1 \right)$ and having slope $m$ is given by:

\(\displaystyle y-y_1=m\left(x-x_1 \right)\)

(2) Difference of squares factorization

\(\displaystyle a^2-b^2=(a+b)(a-b)\)

(3) Discriminant of a quadratic

For the general quadratic $ax^2+bx+c$, the discriminant $\Delta$ is given by:

$\Delta=b^2-4ac$.

The nature of the roots of the quadratic depends in the discriminant. If it is positive , then there are two distinct real roots, if it is zero there is one real (repeated) root, and if it is negative then there are two distinct complex conjugate roots.

If the discriminat is a perfect square, then the roots are rational, otherwise they are irrational.

(4) The quadratic formula

The roots of the general quadratic given in (3) are given by:

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

(5) A consequence of the remainder theorem

If $f(x)$ is divisible by $x-r$, then $f(r)=0$.

(6) The distance formula

The distance $d$ between two points $\left(x_1,y_1 \right),\,\left(x_2,y_2 \right)$ in the plane is given by:

\(\displaystyle d=\sqrt{\left(x_2-x_1 \right)^2+\left(y_2-y_1 \right)^2}\)

(7) The midpoint formula

Given the two points in (6), the midpoint is then:

\(\displaystyle \left(x_M,y_M \right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} \right)\)

Non-Calculator Practice Exam

Question 1

Find the equation of a line passing through the point $(2,6)$ which is parallel to the line with equation $y=2x+1$ expressing it in the form $ax+by+c=0$.

We are given the point $(2,6)$ and the slope $m=2$, since parallel lines have equal slopes. Using the point-slope formula (1), we find that the line may be written as:

\(\displaystyle y-6=2(x-2)\)

\(\displaystyle y-6=2x-4\)

\(\displaystyle y=2x+2\)

Arranging the equation in the desired form, we have:

\(\displaystyle 2x-y+2=0\)

Question 2

Fully factorize each of the following:

a. \(\displaystyle (x+1)^2-25\)

Recognizing that $25$ is the square of $5$, we then have the difference of squares:

\(\displaystyle (x+1)^2-5^2\)

Using the difference of squares formula (2), we then have:

\(\displaystyle ((x+1)+5)((x+1)-5)=(x+6)(x-4)\)

b. \(\displaystyle x^2-4x+1=0\)

Observing that the discriminant (3) $\Delta=(-4)^2-4(1)(1)=12$ is not a perfect square, we know then that the roots are not rational, hence application of the quadratic formula (4) yields:

\(\displaystyle x=\frac{-(-4)\pm\sqrt{12}}{2(1)}=2\pm\sqrt{3}\)

and so we may state:

\(\displaystyle x^2-4x+1=(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))=(x-2-\sqrt{3})(x-2+\sqrt{3})=0\)

c. \(\displaystyle x^3+8x^2+17x+10\), given that \(\displaystyle x+2\) is a factor.

From the given information, we may state:

\(\displaystyle x^3+8x^2+17x+10=(x+2)(x^2+kx+5)=x^3+(k+2)x^2+(5+2k)x+10\)

Equating coefficients, we then find that \(\displaystyle k=6\), hence:

\(\displaystyle x^3+8x^2+17x+10=(x+2)(x^2+6x+5)\)

Now, we may further factor the quadratic factor (since the discriminant is $16$) to get:

\(\displaystyle x^3+8x^2+17x+10=(x+2)(x+1)(x+5)\)

Note, we could also have used the rational roots theorem or polynomial/synthetic division as well. For more information on factoring quadratics, see this topic:

http://www.mathhelpboards.com/f49/factoring-quadratics-3396/

Question 3

For what values of $k$ does \(\displaystyle x^2+kx+4=0\) have one solution?

Recall that a quadratic has a repeated root iff the discriminant (3) is zero. Thus, we require:

\(\displaystyle k^2-4(1)(4)=0\)

\(\displaystyle k^2=16\)

\(\displaystyle k=\pm4\)

Question 4

The cubic \(\displaystyle f(x)=3x^3-(2k+3)x^2+30x-k\) is divisible by \(\displaystyle x-2\). Find the value of $k$.

By the remainder theorem (5), we know:

\(\displaystyle f(2)=0\)

Thus:

\(\displaystyle 3(2)^3-(2k+3)(2)^2+30(2)-k=0\)

\(\displaystyle 3(8)-(2k+3)(4)+30(2)-k=0\)

\(\displaystyle 24-8k-12+60-k=0\)

\(\displaystyle 72=9k\)

\(\displaystyle k=8\)

Question 5

The graph \(\displaystyle y=\frac{a}{x-5}-1\) has a $y$-intercept of 1.

a. SHOW that $a=-10$

We are told that \(\displaystyle y(0)=1\) and so:

\(\displaystyle 1=\frac{a}{0-5}-1\)

\(\displaystyle 2=-\frac{a}{5}\)

\(\displaystyle 10=-a\,\therefore\,a=-10\)

b. State the equations of any asymptotes

There must be a vertical asymptote where division by zero occurs, and so equating the denominator of the first term to zero, we find:

\(\displaystyle x-5=0\,\therefore\,x=5\)

For values of $x$ having unbounded magnitude, i.e., for \(\displaystyle x\to\pm\infty\) the term \(\displaystyle \frac{a}{x-5}\to0\) and to the horizontal asymptote is $y=-1$.

c. Find the exact value of any $x$-intercepts.

Using $a=-10$ and $y=0$, we find:

\(\displaystyle 0=\frac{-10}{x-5}-1\)

\(\displaystyle 1=\frac{10}{5-x}\)

\(\displaystyle 5-x=10\)

\(\displaystyle x=-5\)

Thus, the $y$-intercept is $(-5,0)$.

Question 6

For the triangle in the plane having vertices $A(-1,1),\,B(4,2),\,C(1,4)$

a. Show here that it is an isosceles triangle by proving that the length of $\overline{AC}=\overline{BC}$. Lengths are to be expressed in simplest exact form.

Using the distance formula (6), we find:

\(\displaystyle \overline{AC}=\sqrt{(1+1)^2+(4-1)^2}=\sqrt{4+9}=\sqrt{13}\)

\(\displaystyle \overline{BC}=\sqrt{(1-4)^2+(4-2)^2}=\sqrt{9+3}=\sqrt{13}\)

b. If $M$ is the midpoint of $\overline{AB}$, show that $\overline{CM}$ is perpendicular to $\overline{AB}$.

Using the midpoint formula (7), we find point $M$ is:

\(\displaystyle \left(x_M,y_M \right)=\left(\frac{-1+4}{2},\frac{1+2}{2} \right)=\left(\frac{3}{2},\frac{3}{2} \right)\)

The slope of $\overline{CM}$ is \(\displaystyle \frac{4-\frac{3}{2}}{1-\frac{3}{2}}=-5\)

The slope of $\overline{AB}$ is \(\displaystyle \frac{2-1}{4+1}=\frac{1}{5}\)

Since the produce of these slopes is $-1$, we may conclude they are perpendicular.

CAS Calculator Enabled Practice

Question 1

The midpoint of the line between $(-1,5)$ and $(a,b)$ is $(2,2)$. Find the coordinates of the point $(a,b)$.

Using the midpoint formula (7), we must have:

\(\displaystyle \frac{-1+a}{2}=2\)

\(\displaystyle a-1=4\)

\(\displaystyle a=5\)

\(\displaystyle \frac{5+b}{2}=2\)

\(\displaystyle b+5=4\)

\(\displaystyle b=-1\)

Thus, we have $(a,b)=(5,-1)$.

Question 2

The line \(\displaystyle 5x-y+16=0\) may be written as:

\(\displaystyle y-6=5(x+2)\)

Hence, we know it has gradient $5$ and passes through the point $(-2,6)$.

Question 3

If the two lines $5x-2y+1=0$ and $kx+y-12=0$ are p[erpendicular, then find $k$.

The slope of the first line is \(\displaystyle \frac{5}{2}\) and for the second line is $-k$. We require their product to be $-1$, thus:

\(\displaystyle \frac{5}{2}(-k)=-1\)

\(\displaystyle k=\frac{2}{5}\)

Question 4

An expression for the discriminant of \(\displaystyle mx^2-2mx-4\) is:

\(\displaystyle \Delta=(-2m)^2-4(m)(-4)=4m^2+16m=4m(m+4)\)

Question 5

Find a possible equation for the graph shown:

https://www.physicsforums.com/attachments/870._xfImport

Using the roots, we may state:

\(\displaystyle f(x)=k(x+2)(x-1)(x-3)\) where $k<0$ since \(\displaystyle \lim_{x\to\pm\infty}=\mp\infty\). If we let $k=-1$, then we could write:

\(\displaystyle f(x)=(x+2)(x-1)(3-x)\)
 

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  • #2
Was this for a specific class, or for all of algebra in general?
 
  • #3
Farmtalk said:
Was this for a specific class, or for all of algebra in general?

It was a specific class, but I don't recall the title of the class. The student is in high school, and I know basic calculus was also being taught in this class as well.
 
  • #4
Awesome! I was curious because I have been helping someone practice for the math section of the ACT :D
 
  • #5
=-x^3+4x^2-5x+6

Question 6

Given the graph of f(x)=ax^3+bx^2+cx+d, find the values of $a,\,b,\,c,\,d$.

https://www.physicsforums.com/attachments/871._xfImport

Using the x-intercepts, we may write:

(x+3)(x-2)(x+1)=x^3+2x^2-7x-6

Hence, we may state:

a=1,\,b=2,\,c=-7,\,d=-6

Question 7

Given the graph of f(x)=ax^2+bx+c and f(1)=0, f(2)=3, f(3)=8, find $a,\,b,\,c$.

https://www.physicsforums.com/attachments/872._xfImport

We may write:

f(x)=a(x-1)(x-2)=ax^2-3ax+2a

Since f(3)=8, we have:

9a-9a+2a=8

2a=8

a=4

Now, we may find $b$ and $c$ from the other two conditions:

f(1)=4-3(4)+2(4)=0

f(2)=4-3(8)+2(16)=3

Thus, we may state:

a=4,\,b=-12,\,c=8

Question 8

Given the graph of f(x)=a(x-3)(x+1), find the values of $a$ and $b$.

https://www.physicsforums.com/attachments/873._xfImport

We may write:

f(x)=ax^2-2ax-3a

Since the y-intercept is $-3$, we have:

f(0)=-3

-3a=-3

a=1

Now, we may find $b$ from the fact that the graph passes through the point (2,0):

f(2)=4-4-b=0

b=0

Thus, we may state:

f(x)=(x-3)(x+1)=x^2-2x-3
 

FAQ: Algebra practice exam questions

How can I prepare for an algebra practice exam?

To prepare for an algebra practice exam, it is important to review all of the key concepts and formulas covered in your algebra course. Make sure to practice solving a variety of problems and familiarize yourself with the format and types of questions that will be on the exam. It may also be helpful to create study aids such as flashcards or notes to help you remember important information.

What types of questions can I expect on an algebra practice exam?

An algebra practice exam will typically include a mix of multiple choice, short answer, and problem-solving questions. These may cover topics such as solving equations, factoring, graphing, and working with exponents and polynomials. It is important to review a variety of question types to ensure you are prepared for the exam.

How much time should I spend on each question during an algebra practice exam?

The amount of time you should spend on each question will vary depending on the difficulty of the problem and your own individual pace. It is important to read each question carefully and allocate your time accordingly. If you are struggling with a particular problem, it may be best to move on and come back to it later.

Is it important to show my work when solving algebra practice exam questions?

Yes, it is important to show your work when solving algebra practice exam questions. This not only helps you to check your own answers, but it also allows the instructor to see your thought process and provide partial credit if you make a mistake. Additionally, showing your work can help you to identify where you may have made a mistake if you need to go back and correct an answer.

Can I use a calculator during an algebra practice exam?

This will depend on the specific guidelines set by your instructor. It is important to review any calculator restrictions before taking the exam. In general, most exams will allow the use of a basic scientific calculator, but may prohibit graphing calculators or calculators with advanced features. Make sure to familiarize yourself with your calculator's capabilities and limitations before the exam.

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