Question

Find an equation for the tangent to the curve at the given point.
11) f(x) = 4 x - x + 3, (16, 3)

Answer 1

Use a derivative to find the slope pf the tangent line at that point. Then use point/slope formula to find the equation of the tangent line.

Answer 2

of

Answer 3

The given equation.

Answer 4

f'(x)=3

Answer 5

Yah... hehe. They gave you an easy one for this.

Answer 6

now* what do i do

Answer 7

Point/Slope formula. Remember that from Algebra?

Answer 8

it may be x-x1=slope -x1-x??

Answer 9

y and x.

Answer 10

\(y-y_1=m(x-x_1)\)

Answer 11

so how does the derivative fit into that?

Answer 12

it is m.

Answer 13

so m is 3? or do i need to figure out something

Answer 14

Your \((x_1,y_1)\) is the point given, (16, 3).

When you see the answer, look closely at the original question because it is linear.

Answer 15

so it is y-3=3-16-x

Answer 16

y=-10x

Answer 17

y-3=3(x-16)

Answer 18

so i solve that?

Answer 19

y=6x-45

Answer 20

Yes, but I am kind of wondering about the question you typed in. The point you gave does not fall on that line.

Answer 21

6x?

Answer 22

-45?

Answer 23

ohh ok its missing something

Answer 24

Actually, yah... 45 is right.

Answer 25

it is f(x)=4 square root x -x+3

Answer 26

\[y-3=3(x-16)\implies
y-3=3x-48\implies
y=3x-45\]

OH! \(4 \sqrt{ x} -x+3\)?? That would be a completley different line!

Answer 27

f(x)=4√x-x+3

Answer 28

what to do?

Answer 29

OK... We start over! Wrong derivative because we were working the wrong line.

Answer 30

Slope should be negative with that line.... the root one.

Answer 31

so i find the derivative?

Answer 32

Yep!

Answer 33

f'(x)+1=2/squareroot(x)

Answer 34

You would not move over the 1.

Answer 35

huh?

Answer 36

f'(x)+1=2/√x

Answer 37

You wrote:
\(f'(x)+1=\frac{2}{\sqrt{x}} \)
But it is:
\(f'(x)=\frac{2}{\sqrt{x}} - 1\)

Answer 38

oh ok

Answer 39

Now, this is more like how they normally do these. You need the slope at the point (16,3). So put the 16 into \(f'(x)\) to get the slope at that particular point.

Answer 40

-0.5

Answer 41

\(f'(x)=\frac{2}{\sqrt{x}} - 1\)
\(f'(16)=\frac{2}{\sqrt{16}} - 1\)
\(f'(16)=\frac{2}{4} - 1\)
\(f'(16)=\frac{1}{2} - 1\)
\(f'(16)=-\frac{1}{2}\)
So slope of the tangent line at \((16,3)\) is \(-\frac{1}{2}\)

Yep!

Answer 42

y-3=-0.5(16-x)

Answer 43

Now we are back to \(y-y_1=m(x-x_1)\) Not 16-x, but x-16.

Answer 44

y=1/2x+11

Answer 45

y=2.5x+11

Answer 46

\(y-y_1=m(x-x_1)\)
\(y-3=-\frac{1}{2}(x-16)\)
\(y-3=-\frac{1}{2}x+8\)
\(y=-\frac{1}{2}x+11\)

And that is an equation for the tangent line at that point.

Don;t foget to keep the - on the 1/2 x.

Answer 47

ok thanks. but why isnt that half 2.5

Answer 48

Why would 1/2 suddenly change to something else? The 3 from the other side is added, so it does not distribute to both terms.

Answer 49

\(y-3=-\frac{1}{2}x+8\)
\(y-3+3=-\frac{1}{2}x+8+3\)
\(y=-\frac{1}{2}x+11\)

Answer 50

oh i thought it did

Answer 51

ok thanks

Answer 52

Notice that even though this was a calculus problem, only one step was calculus. The majority was a bunch of algebra. That is how calculus is. Lots of algebra and trig. So if you want to be good at calculus, brush up on the prior stuff. It will help a ton!

Answer 53

ok

Answer 54

Have fun!