Question

Find the equation of the tangent line to the graph f(x)=3x-(24/x²) at the point where x=-2

Our teacher did not even give the correct answer soI don't know where I am at. Can anyone show me how I would go about answering this step by step??

Answer 1

\[\Large f(x)\quad=\quad 3x-\frac{24}{x^2}\]

Answer 2

We're looking for a straight line which is `tangent` to the our curve f(x) at x=-2.
So it will be a line of the form:\[\Large y=mx+b\]Where m, the slope of the line, is given by the derivative of our function f(x) at x=-2.

Answer 3

\[\Large m=f'(-2)\]

Answer 4

Do you understand how to find \(\Large f'(x)\) ?

Answer 5

Yes and no. I am still trying to work on the derivatives and I know I am missing small details when working on there type of problems

Answer 6

Here's a handy rule of exponents.\[\Large \color{royalblue}{\frac{1}{x^n}\quad=\quad x^{-n}}\]

Answer 7

Applying the rule to our function:\[\Large f(x)\quad=\quad 3x-24x^{-2}\]

Answer 8

From this form, we can apply the power rule to each term.

Answer 9

Power rule tells us to do two things:
~Bring the exponent down as a coefficient in front.
~Decrease the exponent by 1.

Answer 10

\[\Large f(x)=3x^1-24x^{-2}\]So our derivative, applying the power rule to each term:\[\Large f'(x)=3(1\cdot x^0)-24(-2\cdot x^{-3})\]

Answer 11

Understand how that works?
The -2 comes down, then we subtract 1 from the exponent, decreasing it to -3.

Answer 12

Yes I am with you so far!

Answer 13

So this will simplify just a bit.\[\Large f'(x)=3+48x^{-3} \qquad\to\qquad f'(x)=3+\frac{48}{x^3}\]

Answer 14

So the `slope` of our tangent line is given by the derivative function.
So what do you get when you plug in x=-2?\[\Large f'(-2)=\;?\]

Answer 15

15? I may be wrong I am just doing it in my head along with studying for my animal morphology test tomorrow lol

Answer 16

sorry I leftout the negative

Answer 17

Mmm I think it's -3ish

Answer 18

Yes yes you are right. That negative I left out made a huge difference! ^^; Thanks so much you really helped me alot on understanding this!

Answer 19

\[\Large m=f'(-2)=-3\]
\[\Large y=mx+b \qquad\to\qquad y=-3x+\color{orangered}{b}\]

Answer 20

So the find the equation of the line tangent to the curve, we need 2 things:
~The slope of the line
~The y-intercept

So we've finished part 1.

Answer 21

To find the y-intercept, we'll need to plug in a point that falls on the line.

Answer 22

I've drawn an example above, hopefully this will help you understand the concept.
The line tangent to the curve `touches` it at the one point.