Question

How do you find the y-intercept(s)?

Answer 1

is the equation in the form y=mx+b?

Answer 2

y = 3x^2 - 10x - 24

Answer 3

its the number without the x. so -24 is the y intercept

Answer 4

How do you find the x intercept?

Answer 5

y = 3x^2 + 10x - 8

Answer 6

look at the graph and see where it intersects on the y axis

Answer 7

-8???

Answer 8

Set "x" =0

Answer 9

first you want to get rid of the exponent and then you want to make sure there is only one x. so you would add the two numbers with the x's. so it would be 9x + (-10x) so the x intercept should be -1x

Answer 10

My choices are a. (2/3,0) and (4,0) b. (-8,0) c. (2/3,0) and (-4,0) d. (8,0)

Answer 11

Y-intercepts are where X = 0. Tthe point is of the form (0, Y). Just plug in x=0 and calculate what y is. It seems the definitions got interchanged above.
X-intercepts are where Y = 0. So we have to solve the equation Y = 0, or 3x^2 + 10x - 8 = 0 for the x-values. The point(s) are of the form (X, 0). This is generally trickier but you just have to solve the quadratic equation in this case. Are you able to do that?

Answer 12

I'm not sure if I'm able, but I'll try.

Answer 13

I'm using ax^2 + bx + c = 0 right?

Answer 14

That is correct. You might try factoring (really the multiple choice gives you the possible factors), or quadratic formula will always work when in doubt.

Answer 15

13x^2 = 8

Answer 16

ugh

Answer 17

As in, you added 8 to both sides:
3x^2 + 10x = 8
And then added together 3x^2 + 10x = 13x^2 ?

Answer 18

3x^2 + 10x = 8
I added the 3x^2 with the 10x to get the 13x^2 = 8

Answer 19

Wrong?

Answer 20

I can't add that because of the exponent right?

Answer 21

Yep.
Adding like this: 3x^2 + 10x = 13x^2
Is kind of like adding like this: 3(2)^2 + 10(2) = 13(2)^2
The left side: 3(2)^2 + 10(2) = 3(4) + 10(2) = 12+20 = 32.
And 13(2)^2 = 13(4) = 52. 32 and 52 are not the same!
So we cannot add 3x^2 and 10x as like-terms, they must be treated differently. :)

Answer 22

Instead, have you used this equation before?
\( x = \dfrac{-b \pm \sqrt{b^2 - 4ac}} {2a} \)
?

Answer 23

no

Answer 24

Ohh, that was the general way to go about it...
how about factoring then? Remember how to factor this form? ax^2 + bx + c, with the number multiplied on x^2?

Answer 25

How do I get rid of the ^2

Answer 26

we basically have to factor the quadratic into this form:
(ax + b)(cx + d) = 0
Then use zero-product to set each piece equal to 0.
ax + b = 0, and cx + d = 0.

Answer 27

OMGod I'm going to fail!

Answer 28

lol

Answer 29

Have you heard of factoring at least? Otherwise, I have another idea that doesn't need any factoring/formula, but it doesn't quite help understand any of the other ways. :P

Answer 30

yes

Answer 31

factoring by grouping

Answer 32

Yep, that's the strategy we'd want ideally. Have any idea of how to approach it? If not, your notes might help, or there are many online resources. :)

Answer 33

Or, I can do one as an example if that helps.

Answer 34

Yes please

Answer 35

Here is a different problem but the same strategy applies. I will use this example:
\( 4x^2 – 19x + 12 = 0 \)

Answer 36

This is in the form ax^2 + bx + c = 0. So we factor by grouping. I start by multiplying the a*c.
4 * 12 = 48
Then finding the factors of 48 that add up to 19. I'd just go through in order.
1, 48 add to 49...
2, 24 add to 27...
3, 16 <-- bingo!
The 19 in the problem is negative, so both factors will be negative.. So let's rewrite:
-19x = -3x - 16x
4x^2 - 3x - 16x + 12 = 0

Answer 37

Notice that we have not changed the value of the problem because -3x - 16x = -19x again once added together. We just changed the writing so it is more convenient for us...

Now the two groups are (4x^2 - 3x) and (-16x + 12).
From the first part, we could pull out an x. x (4x - 3).
From the second part, we could pull out a -4. -4(4x - 3).
So we rewrite: x(4x - 3) - 4(4x - 3) = 0
There is that common factor now, which we should always find when factoring by grouping. Factor out the common (4x - 3) and now we have:
(4x - 3) (x - 4) = 0
This is where zero-product allows us to solve:
4x - 3 = 0 and x - 4 = 0
From there we're just doing linear equations, no more ^2's!

Answer 38

uhhhhh okay
Thank you

Answer 39

Sorry for text walls basically. It is hard to explain otherwise. >.<
But if the process still is unclear, you could try one on your own and we can go over any mistakes. Or take a shot at this problem. And there are also other online resources that explain this stuff:
http://www.purplemath.com/modules/factquad2.htm

Answer 40

My last suggestion is, if all else fails, there is trial and error when you have answer choices. Like, if you wanted to check an answer choice's points you can.
(A) (2/3, 0) and (4, 0)
The equation was y = 3x^2 + 10x - 8. If (A) is true, then these two have to be true:
0 = 3(2/3)^2 + 10(2/3) - 8
0 = 3(4)^2 + 10(4) - 8.
It happens that the first choice does work out to 0 = 0. But the second ends up being 0 = 80, which is not true. So (A) is NOT the correct choice.

Answer 41

Okay thank you!!! I really appreciate it.

Answer 42

You're welcome, and good luck! :)

Answer 43

major headache now

Answer 44

Don't be discouraged if things are not clicking for you right now. At this stage most things are just routines. You learn a process, like factoring (by grouping), and then apply the process to different problems. You become better at a process by simply repeating it, over and over again. It is a boring endeavor, but it takes such an amount of effort in to obtain value out... in anything you want to do.