Solve 2 Calculus Problems: Find Point & Write Equation

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In summary, the two problems are finding the point on a line that lies between two points and solving an equation satisfied by a graph.
  • #1
vee123
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I have two Calculus problems I need explained because I'm studying for a final exam and I forgot how to do these things:

#1: Find the point that lies on the line determined by the points (1,-2) and (-3,1).

The answer is (5,-5), but I don't know why...


#2: Write an equation satisfied by the graph obtained by shifting y=2x-5 three units to the left.

The answer is y=2x+1

I know when an equation is shown moving to the left, it is put in parentheses and a number is added to it (for example y=x^2 becomes y=(x^2+2) when it is moved two places to the left). However, I do not know how to do this problem.

Could someone please help me?
 
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  • #2
vee123 said:
I have two Calculus problems I need explained because I'm studying for a final exam and I forgot how to do these things:

#1: Find the point that lies on the line determined by the points (1,-2) and (-3,1).

The answer is (5,-5), but I don't know why...


#2: Write an equation satisfied by the graph obtained by shifting y=2x-5 three units to the left.

The answer is y=2x+1

I know when an equation is shown moving to the left, it is put in parentheses and a number is added to it (for example y=x^2 becomes y=(x^2+2) when it is moved two places to the left). However, I do not know how to do this problem.

Could someone please help me?

Several things you can do. First off, these aren't calc questions, afaik.

1. Can you write down the equation determined by the two points? If so, then I guess you have choices to plug into the equation to see if they work.

2. The easiest thing to do is to just plot the thing and move it 3 units to the left, and see what you get. But you can also just apply some transforms. In your case, you can shift your function to the left ( f(x) -> f(x+3) ) by applying that transform.

See what you get :)
 
  • #3
Thanks for the reply! I think I'm starting to remember this...

Yeah I know these aren't really calc questions, but they were in the beginning section of my calc. book.

Thanks for your help!
 
  • #4
Anytime :D
 
  • #5
Okay, I'm still having trouble with the second question. I know I may sound really stupid, but I tried to solve it by doing f(x)=(2x-5+3), and did not get the correct answer. What am I doing incorrectly?
 
  • #6
Consider rewriting f(x) = 2x-5 as f(x) = 2(x-5/2).. and see if you get anything from there...
 
  • #7
Now I understand! I was having a bit of a mental block at hte moment... Thank you very much! :)
 

1. How do I find the point of intersection for two curves?

To find the point of intersection for two curves, you will need to set the equations of the two curves equal to each other. Then, solve for the variable to determine the x-coordinate. Once you have the x-coordinate, plug it back into either of the original equations to find the corresponding y-coordinate.

2. What is the equation for a line passing through two given points?

The equation for a line passing through two given points (x1,y1) and (x2,y2) can be found using the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. The slope, m, can be calculated by taking the difference in the y-coordinates divided by the difference in the x-coordinates, which is (y2-y1)/(x2-x1). Once you have the slope, plug it into the equation along with one of the given points to solve for b. This will give you the equation for the line.

3. Can I use calculus to solve for the point of inflection?

Yes, you can use calculus to find the point of inflection for a curve. The point of inflection is where the second derivative of the curve changes sign, going from positive to negative or vice versa. To find this point, take the second derivative of the curve and set it equal to 0. Then, solve for the variable to determine the x-coordinate. Plug this x-coordinate back into the original equation to find the corresponding y-coordinate.

4. How do I know if a point is a critical point?

A critical point is a point where the derivative of the curve is equal to 0, or where the derivative does not exist. To determine if a point is a critical point, take the first derivative of the curve and set it equal to 0. Then, solve for the variable to determine the x-coordinate. Plug this x-coordinate back into the original equation to find the corresponding y-coordinate. If the derivative does not exist at this point, it is also considered a critical point.

5. Can I use calculus to find the maximum or minimum value of a curve?

Yes, you can use calculus to find the maximum or minimum value of a curve. The maximum or minimum value occurs at the point where the derivative is equal to 0 or does not exist. To find this point, take the first derivative of the curve and set it equal to 0. Then, solve for the variable to determine the x-coordinate. Plug this x-coordinate back into the original equation to find the corresponding y-coordinate. You can also use the second derivative test to determine if the point is a maximum or minimum point.

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