# horizontal or vertical?

• Sep 4th 2009, 07:44 PM
pepperroni
horizontal or vertical?
The graph of y=5X^2-5 is the graph of y=X^2-1 stretched vertically or horizontally?

I get that it streches by 5 but how do i tell if its vertical or horizontial?

also

What are the x and y intercepts of F(x)=4xsquared-1?

i got +or-1/8 for the x intercept but im sure its wrong. I thought i got it in class but i guess not. Why is math pure 30 required for college :((Headbang)
• Sep 4th 2009, 08:40 PM
aidan
Quote:

Originally Posted by pepperroni
The graph of y=5X^2-5 is the graph of y=X^2-1 stretched vertically or horizontally?

I get that it streches by 5 but how do i tell if its vertical or horizontial?

also
....

1) Exclude the constants.
2) Plot/sketch/graph both functions
3) What would you have to do to get y=x^2 to overlay y=5x^2?
Would you stretch it horizontally or vertically?
• Sep 6th 2009, 07:59 PM
pepperroni
thanks i dont know why i didnt think of that... but is the only way to find if its vert or horizontial is by comparing? this is just me being curious now.
• Sep 7th 2009, 04:14 AM
HallsofIvy
Any thing done before the "main function" changes x (and so affects the graph horizontally), anything done after the "main function" changes y (and so affects the graph vertically).

Here your "main function" is given as $\displaystyle y= x^2- 1$. $\displaystyle 5x^2- 5= 5(x^2- 1)$ multiplies by 5 after doing $\displaystyle x^2- 1$ and so affects the graph vertically.

The x-intercepts are where the graph "intercepts" the x-axis and so y= 0. $\displaystyle y= 4x^2- 1= 0$ is the same as $\displaystyle 4x^2= 1$ or $\displaystyle x^2= \frac{1}{4}$. The x-intercepts are at x= $\displaystyle \frac{1}{2}$ and $\displaystyle x= -\frac{1}{2}$. Some people refer to the "x-intercepts" as just the x value ($\displaystyle \frac{1}{2}$ and $\displaystyle -\frac{1}{2}$) and some as the points ($\displaystyle \left(\frac{1}{2}, 0\right)$ and $\displaystyle \left(-\frac{1}{2}, 0\right)$. Check with your teacher or textbook to see which convention is used in your class.

The y-intercepts are where the graph "intercepts" the y-axis and so x= 0. Just put x= 0 into $\displaystyle y= 4x^2- 1$ to get y= -1. Again, the y-intercept is either the value y= -1 or the point (0, -1) depending on the convention used.

I have no idea what "math pure 30" is but (liberal arts) colleges want you to be educated in a broad range of subjects. You will very likely be required to take at least one mathematics course in college.
• Sep 7th 2009, 08:52 AM
pepperroni
Quote:

Originally Posted by HallsofIvy
Any thing done before the "main function" changes x (and so affects the graph horizontally), anything done after the "main function" changes y (and so affects the graph vertically).

Here your "main function" is given as $\displaystyle y= x^2- 1$. $\displaystyle 5x^2- 5= 5(x^2- 1)$ multiplies by 5 after doing $\displaystyle x^2- 1$ and so affects the graph vertically.

The x-intercepts are where the graph "intercepts" the x-axis and so y= 0. $\displaystyle y= 4x^2- 1= 0$ is the same as $\displaystyle 4x^2= 1$ or $\displaystyle x^2= \frac{1}{4}$. The x-intercepts are at x= $\displaystyle \frac{1}{2}$ and $\displaystyle x= -\frac{1}{2}$. Some people refer to the "x-intercepts" as just the x value ($\displaystyle \frac{1}{2}$ and $\displaystyle -\frac{1}{2}$) and some as the points ($\displaystyle \left(\frac{1}{2}, 0\right)$ and $\displaystyle \left(-\frac{1}{2}, 0\right)$. Check with your teacher or textbook to see which convention is used in your class.

The y-intercepts are where the graph "intercepts" the y-axis and so x= 0. Just put x= 0 into $\displaystyle y= 4x^2- 1$ to get y= -1. Again, the y-intercept is either the value y= -1 or the point (0, -1) depending on the convention used.

I have no idea what "math pure 30" is but (liberal arts) colleges want you to be educated in a broad range of subjects. You will very likely be required to take at least one mathematics course in college.

How did you go from 1/4 and get 1/2 and -1/2 thats were im lost i get the idea and i got that far myslef but then im not sure how to get 1/4 to the two answers you got..... sorry to be a bother... im horrible at math, but I have to make it through this class. Pure 30 is our high schools hardest math class... our teacher says the average for the class is 60 and many will not pass XD
• Sep 7th 2009, 09:18 AM
skeeter
Quote:

Originally Posted by pepperroni
How did you go from 1/4 and get 1/2 and -1/2 thats were im lost i get the idea and i got that far myslef but then im not sure how to get 1/4 to the two answers you got..... sorry to be a bother... im horrible at math, but I have to make it through this class. Pure 30 is our high schools hardest math class... our teacher says the average for the class is 60 and many will not pass XD

4x^2 - 1 = 0

(2x - 1)(2x + 1) = 0

set each factor equal to 0 and solve for x.