Solving Exercise 5 Algebra II Honors

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darshanpatel
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I don't get the wording of the problem. What does it mean find the equation?

Problem: Find the equation of the secant line through the points (x, f(x)) and (x2, f(x2)) for parts (a)-(e) in Exercise 5.

There is no work because I don't know where to begin.

Just as an example, (a) in Exercise 5 is x=2 and x2=3...
What do I do?

They also give me m(sec)= f(x2)-f(x)/x2-x
 
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darshanpatel said:
I don't get the wording of the problem. What does it mean find the equation?

Problem: Find the equation of the secant line through the points (x, f(x)) and (x2, f(x2)) for parts (a)-(e) in Exercise 5.

There is no work because I don't know where to begin.

Just as an example, (a) in Exercise 5 is x=2 and x2=3...
What do I do?

They also give me m(sec)= f(x2)-f(x)/x2-x

The equation of a line is often of the form
y = mx + b

for instance, y = 3x - 7​
 
I know it is y=mx+b but how would i find that? Can you please show using the example numbers?
 
Yes for exercise 5 it said consider the function given by f(x)=sqrt(x-1)
 
what do you mean?
 
oh, f(2)=1 and f(3)=sqrt2
 
I tried putting it into point-slope form and got y-1=(sqrt2-1)(x-2), is that a right start?

Reduced that down to y=(sqrt2x -x)-2sqrt2 +3
 
Last edited:
SammyS u still want to help me?
 
darshanpatel said:
I tried putting it into point-slope form and got y-1=(sqrt2-1)(x-2), is that a right start?
Yes
Reduced that down to y=(sqrt2x -x)-2sqrt2 +3
Right, though it is usually better to write it with just one x,
viz., y = (sqrt2 - 1)x -2sqrt2 + 3

But you are not finished yet. The final step is to check that this equation fits your initial data, to avoid the embarrassment of scoring some red crosses when your work is marked.

When x=2, does this produce a y value of 1?
when x=3, does this give y = sqrt2?

If this all tallies, then it must be right. :smile:
 
darshanpatel said:
I tried putting it into point-slope form and got y-1=(sqrt2-1)(x-2), is that a right start?

Reduced that down to y=(sqrt2x -x)-2sqrt2 +3
This -- sqrt2x -- is terrible notation because it is ambiguous. Does it mean [itex]\sqrt{2x}[/itex] or does it mean [itex]\sqrt{2}\cdot x[/itex]? Because this expression came from (=(√2-1)(x-2), you apparently intend for what you wrote to mean [itex]\sqrt{2}\cdot x[/itex]. A better way to write that is to put x in front of the radical, as x√2, which is clear and unambiguous.
 
Thank you, sorry, i see it written a lot like sqrt2 or what ever, but how do you do the symbols?
 
darshanpatel said:
Thank you, sorry, i see it written a lot like sqrt2 or what ever, but how do you do the symbols?
To use the typesetting fonts and symbols, you have to invest time in learning Latex-family formatting. This site makes it as effortless as possible: http://www.codecogs.com/latex/eqneditor.php

You can construct your itex formatting on that site, then cut and paste it into your posts.