Finding equation for a line?

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February 1st 2010, 07:56 PM
nox16

Finding equation for a line?

I don't even know to to start on this problem. It doesn't even give you any coordinates at all so how am I suppose to find the equation for this line?

http://i45.tinypic.com/1y2zpe.jpg
February 1st 2010, 08:01 PM
pickslides

Well the equation of a line in this case can be given by $y = mx+c$

They have told you that $m= 4$ so $y = 4x+c$

They have also told you the line shares a point with the parabola. This point is $(2, 2^2) = (2,4)$ so

$4 = 4\times 2+c$

Now solve for $c$ and your done.
February 1st 2010, 08:06 PM
snaes

Use the form $y=mx+b$

m is slope
b is y intercept

m=4 (given)

Put in the equation:
$y=4x+b$
We are given in the graph a point x=2 there is an interception with the graph $y=x^2$
so $y=(2)^2$ or $y=4$ at that point.

Now you know the point (2,4) is on the graph.

put in 2 for x and 4 for y in the $y=4x+b$ and solve for b:

$y=mx+b$
$4=4*2+b$
$b=1/2$

now you are done!
$y=4x+.5$
February 1st 2010, 08:06 PM
nox16

Nevermind snaes answered my question. Thank you so much for helping me with this question!

I just got the answer and it is y = 4x -4. I think you kind of got it mixed up snaes.
February 1st 2010, 08:08 PM
pickslides

correct.
February 1st 2010, 08:13 PM
bigwave

also another way to get

$y = mx+c$

is to use the point slope form
$y-y_1 = m(x - x_1)$

so using point x = 2 and since y=2^2 = 4 with have the point (2,4)

then

$y-4 = 4(x-2)$
$y = 4x - 8 +4$
$y = 4x -4$
February 1st 2010, 08:16 PM
pickslides

Quote:

Originally Posted by snaes

$y=mx+b$
$4=4*2+b$
$b=1/2$

now you are done!
$y=4x+.5$

This is not correct.
February 1st 2010, 08:26 PM
nox16

Thank you so much for your help guys I really appreciate it. Now I am stuck on this problem. How would I find out the slope of it? I already plugged in the x and y intercepts it gave me into the equation y = x^2 + 1 and got the coordinate (square root of 7, 5). I'm thinking the top of the triangle is 10 and the right side might be 1 but I'm not quite sure. So that makes it a rise over run of 1/10. Is that the correct slope?

http://i48.tinypic.com/n3of1d.jpg
February 1st 2010, 08:34 PM
bigwave

you have to do the rise/run based on the 2 points

first find the (x,y) of the 2 intersecting points

then use $m = \frac{y-y_1}{x-x_1}$ for the slope

as you have found

$Q(2,5)$ and $P(\sqrt{7},8)$

then

$m = \frac{8-5}{\sqrt{7}-2}$

then just simplify

I got $m=\sqrt{7} + 2$ but double check this
February 1st 2010, 09:04 PM
nox16

I don't think that is the right answer. Here's what it says in the back of the book: http://i46.tinypic.com/2zokck9.jpg

At this time I got the slope 3/sqrt{7} - 2 and is trying to plug in (2,5) into the equation y = mx + b. But I get stuck at this point 5 = 3/sqrt{7} - 2 (2) + b
February 2nd 2010, 12:15 PM
pickslides

Quote:

Originally Posted by bigwave

$m = \frac{8-5}{\sqrt{7}-2}$

then just simplify

I got $m=\sqrt{7} + 2$ but double check this

Ray (Bigwave) has rationalised the denominator to get $m=\sqrt{7} + 2$

i.e. $m = \frac{3}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}=\sqrt{7} + 2$

It is not required in this case.
February 2nd 2010, 03:50 PM
nox16

Thanks so much for helping me with this problem.