Equation of tagent line to the graph

[Tjvelcro]

Homework Statement

a. Find the equation of the tangent line to the graph of y=g(x) at x = 5 if g(5) = -3 and g'(5) = 4

b. If the tangent line to y = f(x) at (4,3) passes through the point (0,2), find f(4) and f'(4)

Homework Equations

y=mx =b
f(a+h)-f(a)
h
y – y1 = m(x – x1)

The Attempt at a Solution

So in a. I'm given a point (5, -3) but I'm not sure if we can find the equation right away or find the slope first. I also don't know what g'(5) = 4 has to do with the problem. Should I sketch them?

For b. I'm not even sure what I'm trying to find, a value? equation?

 

[Mark44]

Homework Statement

a. Find the equation of the tangent line to the graph of y=g(x) at x = 5 if g(x) = -3 and g'(5) = 4

Instead of g(x) = -3, you mean g(5) = 3. g(x) = -3 means that for any x, the y value is -3.

b. If the tangent line to y = f(x) at (4,3) passes through the point (0,2), find f(4) and f'(4)

Homework Equations

y=mx =b
f(a+h)-f(a)
h
y – y1 = m(x – x1)

The Attempt at a Solution

So in a. I'm given a point (5, -3) but I'm not sure if we can find the equation right away or find the slope first. I also don't know what g'(5) = 4 has to do with the problem. Should I sketch them?

You have a point -- (5, -3) -- that's on the curve and on the tangent line, and you have the derivative of g at 5 -- g'(5) -- that is the slope of the tangent line. Given a point on a line and its slope, what is the equation of the line?

For b. I'm not even sure what I'm trying to find, a value? equation?

For b, you are suppose to find two values, f(4) and f'(4). That's why it says "find f(4) and f'(4)."

You are apparently missing the connection between the value of the derivative at a point and the slope of the line that is tangent to the curve at that point.

For this problem you are given two points on the tangent line (hint: find the slope of this line) and the point of tangency -- (4, 3). Finding f(4) is really a no-brainer.

 

[Tjvelcro]

a. so I use y – y1 = m(x – x1)
sub in (5, -3) for the point and 4 for the slope?
Solve for y.
So y-(-3)=4(x-5)
y=4x-23

b. I would use m=(y2-y1)/(x2-x1) to find the slope as you suggest.
m=(3-2)/(4-0)
m=1/4

Now I need to find f(4) and f'(4)
I need to find the equation first as I did in part a.
so I use y – y1 = m(x – x1)
sub in (4,3) for the point and 1/4 for the slope?
So y-(3)=1/4(x-4)
y=(1/4)x+2

Now I can solve for f(4) but is f'(4) any different?

y=(1/4)(4)+2
y=3

 

[Mark44]

a. so I use y – y1 = m(x – x1)
sub in (5, -3) for the point and 4 for the slope?
Solve for y.
So y-(-3)=4(x-5)
y=4x-23

Correct.

b. I would use m=(y2-y1)/(x2-x1) to find the slope as you suggest.
m=(3-2)/(4-0)
m=1/4

Correct.

Now I need to find f(4) and f'(4)

Don't overthink these. (4, 3) is a point on the graph of f, so what is f(4)?

For f'(4) you are still missing the connection between the 1) slope of the tangent line to a function at a point (a, f(a)) on the graph and 2) the value of f'(a). For this problem, once you have found the slope of the tangent line (m = 1/4), you DON'T NEED TO DO ANY OTHER CALCULATIONS!

In particular, you ARE NOT ASKED to find the equation of the tangent line - just f(4) and f'(4).

I need to find the equation first as I did in part a.
so I use y – y1 = m(x – x1)
sub in (4,3) for the point and 1/4 for the slope?
So y-(3)=1/4(x-4)
y=(1/4)x+2

Now I can solve for f(4) but is f'(4) any different?

y=(1/4)(4)+2
y=3

 

[Tjvelcro]

So f(4)=3 as given in the question...
f'(4)=1/4 since 1/4 is the slope of the tangent line at point (4,3)?

Thanks for the help! Test this Thursday!

 

[Mark44]

So f(4)=3 as given in the question...
f'(4)=1/4 since 1/4 is the slope of the tangent line at point (4,3)?

Right.