Homework Help

[ambright4life]

I have answers to the questions, so can someone help me with letting me know if they're correct or not? I would greatly appreciate it!

Consider the graph of the function f(x)=x^2-x-12. (a) Find the equation of the secant line joining the points (-2, -6) and (4, 0). (b) Use the Mean Value Theorem to determine a point c in the interval (-2, 4) such that the tangent line at c is parallel to the secant line. (c) Find the equation of the tangent line through c.

a) y=x-4
b) f'(c)=1
c)y=x-13

Thank you in advance!

 

[DrPhil]

I have answers to the questions, so can someone help me with letting me know if they're correct or not? I would greatly appreciate it!

Consider the graph of the function f(x)=x^2-x-12. (a) Find the equation of the secant line joining the points (-2, -6) and (4, 0). (b) Use the Mean Value Theorem to determine a point c in the interval (-2, 4) such that the tangent line at c is parallel to the secant line. (c) Find the equation of the tangent line through c.

a) y=x-4
b) f'(c)=1
c)y=x-13

Thank you in advance!

Fine!

The way the question is worded, should (b) be c=(1,-12) instead of f'(c)=1 ? [both are true]

 

[Deleted member 4993]

I have answers to the questions, so can someone help me with letting me know if they're correct or not? I would greatly appreciate it!

Consider the graph of the function f(x)=x^2-x-12. (a) Find the equation of the secant line joining the points (-2, -6) and (4, 0). (b) Use the Mean Value Theorem to determine a point c in the interval (-2, 4) such that the tangent line at c is parallel to the secant line. (c) Find the equation of the tangent line through c.

a) y=x-4
b) f'(c)=1
c)y=x-13

Thank you in advance!

Looks good except.....

Part (b) is asking you to determine the point (c) where f'(c) = 1

 

[ambright4life]

f'(c)=1
2c-1=1
c=1

So would the answer for part b would be c=1 right?

 

[Bob Brown MSEE]

f'(c)=1
2c-1=1
c=1

So would the answer for part b would be c=1 right?

Right! Infact no matter what two points (x1,y1) and (x2,y2) are chosen for the secant -- c=Avg(x1, x2)

This is true for any quadratic polynomial of the form y = xx + ax + b.

 

[DrPhil]

f'(c)=1
2c-1=1
c=1

So would the answer for part b would be c=1 right?

The value of x at point c is 1.

Technically, a "point" is (x,f(x)), so a purist would prefer for you to say c = (1, -12). You won't lose any points if you say (1,-12). For me it's a matter of style, though when you say c=1, I know you have found the correct answer. What does your teacher want?

 

[ambright4life]

I don't know what my teacher want. She just want us to answer the question, but thank you guys so much! You really helped me a lot!