Algebra II help: line equations, polynomial division, etc.

[Lydia823]

I have some questions regarding my math summer homework. The first one is

1) Complete the following linear function table

  x        y

  1/2     4 1/2

1 1/2    12 1/2

3 1/2       ?

3 3/4    30 1/2   

4 1/2    36 1/2

2) Write an equation for the line that passes through the points (8, ½) and (4, ¼)

3) Given the cooridinate points A(2, 4), C(3, 6), and F (4, 8), which equation proves that point C lies on the line AF?

A. y=2x+12

B. y=2x

C. y=x-12

D. y= -2+12

4) The second of two numbers is equal to three times the first. The second number decreased by eight is equal to the first number multiplied by negative two. Find the numbers.

5) Divide (2x squared – 14x – 88) by (x – 11)

I had forty questions to do. These are the only five I did not know how to do. Could you please explain them and tell me how to get the answer and what the answer might be?

Thank you very much,

Lydia

 

[galactus]

I'll help with the 5th one:

It's matter of factoring the numerator and cancelling.

Factor \(\displaystyle \L\\2x^{2}-14x-88\)

\(\displaystyle \L\\2(x^{2}-7x-44)\)

Use what's in the parentheses and ask yourself, what two numbers when added equal -7 and when multiplied equal -44

Would it be -11 and 4?. (-11)(4)=-44, -11+4=-7

\(\displaystyle \L\\x^{2}-11x+4x-44\)

\(\displaystyle \L\\(x^{2}-11x)+(4x-44)\)

\(\displaystyle \L\\x(x-11)+4(x-11)\)

\(\displaystyle \L\\(x+4)(x-11)\)

Don't forget your 2:

\(\displaystyle \L\\2(x+4)(x-11)\)

Now, you have:

\(\displaystyle \L\\\frac{2(x+4)(\sout{x-11})}{\sout{x-11}}\)

2(x+4)

 

[jonboy]

Re: Algebra II help

Hello Lynia

[2] The second one is Write an equation for the line that passes through the points (8, ½) and (4, ¼)

[4]The fourth is The second of two numbers is equal to three times the first. The second number decreased by eight is equal to the first number multiplied by negative two. Find the numbers.

[2] For [2] can you use slope-intercept form on this (or is that obsolete in Algebra 2)?

If not then we will use: \(\displaystyle \L y=mx+b\) in which \(\displaystyle \L m\) is the slope and constant \(\displaystyle \L b\) is the y-intercept since we are given two points.

\(\displaystyle \L m=\frac{\Delta y}{\Delta x}=\frac{\frac{1}{2} -\frac{1}{4}}{8-4}=\frac{\frac{1}{4}}{4}=\frac{1}{16}\)

Fill in one of our points to find \(\displaystyle \L b\): \(\displaystyle \L y=\frac{1}{16}x+b\)

\(\displaystyle \L \frac{1}{2}=\frac{1}{16}(8)+b\)

\(\displaystyle \L \frac{1}{2}=\frac{1}{2}+b\)

\(\displaystyle \L 0=b\)

\(\displaystyle \L y=\frac{1}{16}x\)

[4] We will call \(\displaystyle \L x\) the first number and \(\displaystyle \L y\) the second number.

Equation 1-The second of two numbers is equal to three times the first=\(\displaystyle \L y=3x\)

Equation 2-The second number decreased by eight is equal to the first number multiplied by negative two. \(\displaystyle \L y-8=-2x\)

The easiest way I see of doing this through subsititution is by finding \(\displaystyle \L x\) in terms of \(\displaystyle \L y\)

Given: \(\displaystyle \L y=3x\)

Division Propery: \(\displaystyle \L x=\frac{y}{3}\)

So we subsitute that in our second equation: \(\displaystyle \L y-8=-2(\frac{y}{3})\)

Also can be written as: \(\displaystyle \L y-8=\frac{-2}{1}(\frac{y}{3})\)

Subsitution Property: \(\displaystyle \L y-8=\frac{-2y}{3}\)

Subtraction Propery: \(\displaystyle \L -8= -1\frac{2}{3}y\)

Division Property: \(\displaystyle \L y=4\frac{4}{5}\)

Fill that in our first equation to find \(\displaystyle \L x\):

\(\displaystyle \L y=3x\to4\frac{4}{5}=3x\to x=1\frac{3}{5}\)

Check: \(\displaystyle \L 4\frac{4}{5}=3(1\frac{3}{5})\)

\(\displaystyle \L 4\frac{4}{5}=4\frac{4}{5}\)

\(\displaystyle \L y-8=-2x\)

\(\displaystyle \L 4\frac{4}{5}-8=-2(1\frac{3}{5})\)

\(\displaystyle \L -3\frac{1}{5}= -3\frac{1}{5}\)

 

[jonboy]

Hello again

[3]The third is given the cooridinate points A(2,4), C(3,6), and F (4,8), which equation proves that point C lies on the line AF?

Here is what the line would look like graphed:

Ok since we are trying to prove F(4,8) is on the line fill in 4 in each equation (since it is an x-coordinate fill it in for x of course) to see if you get 8 (y-coordinate) or just find the answer yourself by using slope-intercept form: \(\displaystyle \L y=mx+b\) since all your answers are in that form. See my previous post for an example of using slope-intercept form.

 

[jonboy]

Re: Algebra II help

I have some questions regarding my math summer homework. The first one is Complete the following linear function table

X y

½ 4 1/2

1 ½ 12 1/2

3 ½ ?

3 ¾ 30 1/2

4 ½ 36 ½

Well since this is a line you can use slope intercept form to create a formula by using \(\displaystyle \L y=mx+b\) and plug in \(\displaystyle 3\frac{1}{2}\) for x.

Now you are all set to solve all 5 of these problems.