Finding the zeros and intercepts---help / check answer?

**tiar** · September 15th 2008, 03:34 PM

Hi, again I am rusty at math (I haven't done it in two years). There are no solutions in my book for my homework, so could someone double check that I have the right idea?

1) Find the zeros of the given function f.
$f(x)=x^3-x^2-2x$
$f(x)=x(x^2-x-2)$
$f(x)=x(x-2)(x+1)$

So, are the zeros x=2, x=-1, and x=0?
One more...

2) Find the x- and y-intercepts, if any, of the graph of the given function f. Do not graph.

$f(x)=(2x-3)(x^2+8x+16)$
$f(x)=2x^3+16x^2+32x-3x^2-24x+13$
$f(x)=2x^3+13x^2+8x+13$
$f(x)=x(2x^2+13x+21)$
$f(x)=x(2x+7)(x+3)$

For the X-intercepts I got...
(-7/2,0), (-3,0), (0,0)

How do I find the y-intercepts then? Do I plug in zero for "x"? If I do, wouldn't the y-intercept be just (0,0)?

Thank you so much in advanced!

**11rdc11** · September 15th 2008, 03:50 PM

yes plug in 0 for x to find y intercept. The y intercept is (0,13)

**tiar** · September 15th 2008, 04:03 PM

Originally Posted by 11rdc11

yes plug in 0 for x to find y intercept. The y intercept is (0,13)

Thank you so much! I highly appreciate it. Do the other questions look alright?

**icemanfan** · September 15th 2008, 04:10 PM

Just to clarify, these two functions are not the same:
$f(x)=2x^3+13x^2+8x+13$

$f(x)=x(2x+7)(x+3)$

The first has a y-intercept of (0, 13) and the second has a y-intercept of (0, 21). Did you try to find x-intercepts for the first equation at all?

**tiar** · September 15th 2008, 04:15 PM

Originally Posted by icemanfan

Just to clarify, these two functions are not the same:
$f(x)=2x^3+13x^2+8x+13$

$f(x)=x(2x+7)(x+3)$

The first has a y-intercept of (0, 13) and the second has a y-intercept of (0, 21). Did you try to find x-intercepts for the first equation at all?

Yes, I did (first post). I was unsure if it was correct.

(-7/2, 0), (-3, 0), (0, 0)

So you must always find the y-intercept of both functions, not just the first one? Thanks in advanced.

**11rdc11** · September 15th 2008, 04:51 PM

You made your mistake right here, this is not true

$2x^3+13x^2+8x+13 = x(2x^2+13x+21)$

The people who made the problem gave

$<br /> f(x)=(2x-3)(x^2+8x+16)<br />$

instead of distributing it out

$2x^3+13x^2+8x+13$

so you could find your x intercepts easier.

$0=2x-3$

$0 = x^2+8x+16$

**tiar** · September 15th 2008, 06:41 PM

Ah! Thank you so much...I totally forgot about this. So the correct way would be:

$0=2x-3$

$\frac 32 = x$

$x^2+8x+16$
$(x+4)(x+4)$
$x=-4$

So the x-intercepts are ( $\frac 32$ , 0) and (-4, 0)

**11rdc11** · September 15th 2008, 06:49 PM

Yep

**tiar** · September 15th 2008, 07:03 PM

Yay! Thank you! I'm so glad someone can point out my silly mistakes, haha.
The homework is going alright so far, but I'm a little uneasy that I have the correct way of doing this.

Find an equation of the line that satisfies the given conditions:
through the origin parallel to the line through (1,0) and (-2,6)

So my way of thinking was: find the slope since it is the same as the parallel line, and then the origin is at (0,0), correct? Thus...

$m= \frac {6-0} {-2-1}$

$m=-2$

plug it into the point-slope equation...

$y-0^2=m(x-0^2)$

therefore the equation would be...(I hope)

$y=-2x$

**11rdc11** · September 15th 2008, 07:06 PM

$y_2 - y_1 = -2(x_2 - x_1)$

$y - 0 = -2(x - 1)$

Use either of your original points

**tiar** · September 15th 2008, 07:15 PM

Originally Posted by 11rdc11

$y_2 - y_1 = -2(x_2 - x_1)$

$y - 0 = -2(x - 1)$

Use either of your original points

So if it says "through the origin parallel to the line through (1,0) and (-2,6)"...we can use those points which are parallel to the line it's talking about? Texts are so awkwardly worded sometimes, it makes my head spin.

So, the equation would be...
$y-1^2=-2x(x-0^2)$
$y-1=-2x$

**11rdc11** · September 15th 2008, 07:22 PM

Oops I didn't read the orgin part so your answer is correct

$y = -2x$

Check question, why are you squaring here

**tiar** · September 15th 2008, 07:31 PM

Originally Posted by 11rdc11

Oops I didn't read the orgin part so your answer is correct

$y = -2x$

Check question, why are you squaring here

Oh okay! Thanks!
HAHA yes that's what I get for my bad sleeping habits. That's meant to be y2 and x2...not squared. Sorry, I glanced over the formula super fast. Ignore that...

Anyway, I know you are sick of answering math problems. Maybe someone else will be able to help. I really hope I can get back into math in the next few weeks so I can figure things out on my own...

I just need explaining on this one:
"If C is an arbitrary real constant, an equation such as 2x - 3y = C is said to define a family of lines. Choose four different values of C and plot the corresponding lines on the same coordinate axes. What is true about hte lines that are members of this family?"

Do I plug in numbers for C? If I do, how do I solve that?
I was thinking to plug in numbers for x and y but I do not think it's asking that...

**11rdc11** · September 15th 2008, 07:43 PM

$2x - 3y = C$

$-3y = -2x + C$

$y = \frac{2x}{3} -\frac{C}{3}$

Ok now put in whatever value for C

The only thing that is happening is the graph is moving up or down. It is easier to see what I'm trying to explain. Try plooting it and you will see what I mean.

**tiar** · September 15th 2008, 07:57 PM

Thank you so much! That makes way more sense. I will play around with that and see what I get. *must review notes...*

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