0
$\begingroup$

I have a math problem that I am a little "iffy" on.

1: Find the slope of the curve at point $P(1,4)$ and find the equation of the tangent line at P.

$y=5-x^2 P(1,4)$.

Now to find the slope, is $(y_2-y_1)/(x_2-x_1)$, but to do this, I need to know point Q. But I am not 100% sure I found Q correctly. I got $Q(1+h),(2+h)^2$ but I am a little stuck there.

So

$(2+h)^2-1^2/h$ would be the equation to find the slope? Once I know the slope I think I can just plug and chug point slope formula for the tangent line using P and the slope.

Thank you

$\endgroup$
  • 2
    $\begingroup$ Wouldn't the most effective way to do this just be to find the derivative at point P, and plug that as the slope in the equation $y - y_0 = (x-x_0)m$? $\endgroup$ – Maroon Feb 13 '14 at 1:37
  • 3
    $\begingroup$ The question statement does not imply you have to differentiate everything from scratch. You should be able to just use $\frac{d}{dx} x^n = nx^{n-1}$ to calculate the slope. $\endgroup$ – Felipe Jacob Feb 13 '14 at 1:37
1
$\begingroup$

To find the slope of the curve, or the gradient you can differentiate the function and plug in the $x$ value of the point into the derivative and that will yield the gradient at that point. So you have $y=5-x^2$ at the point $(1, 4)$. $$y'=-2x$$ The gradient of the curve, $y=5-x^2$ at $(1,4)$ is: $$y'(1)=-2(1)=-2$$

For the equation of the tangent line, you have the gradient $(-2)$, a point on the line $(1, 4)$, and you know that the equation of a line is $y=mx+c$. You can plug in the values that you know to find the value of $c$ and then you'll have the equation of the tangent line.

$\endgroup$
  • $\begingroup$ May I ask another question in this thread? I have to find the limit of lim x->0 of √(5x+4)-2/x and even with manipulation, I still got that it does not exist, since 7/0=undefined. That was after multiplying by √(5x+4)-2 for top and bottom $\endgroup$ – Terry Feb 13 '14 at 1:58
  • $\begingroup$ It is not customary to ask two questions in the same thread as you put it, but I'll help you since I see your problem. Your mistake is that you multiplied it by itself. What you should do is multiply it by it conjugate, $\sqrt{5x+4}+2$. So: $$\lim_{x\to 0} \frac{\sqrt{5x+4}-2}{x}\cdot\frac{\sqrt{5x+4}+2}{\sqrt{5x+4}+2}$$ will give you what you need. @Terry $\endgroup$ – Zhoe Feb 13 '14 at 2:04
  • $\begingroup$ ah thank you. was my y=-2x+6 ok? $\endgroup$ – Terry Feb 13 '14 at 2:18
  • $\begingroup$ Yes, sorry I didn't say. That's correct. $\endgroup$ – Zhoe Feb 13 '14 at 2:27
  • $\begingroup$ Thank you very much,i forgot its the conjugate, saved me on 2 other problems. Thanks so much $\endgroup$ – Terry Feb 13 '14 at 2:29
0
$\begingroup$

Since $$y=5-x^2\\ \frac{dy}{dx}=-2x$$ This gives for the tangent line: $$y-y_1=\frac{dy}{dx}(x-x_1)\\ =y-y_1=2x(x_1-x)$$ At $P=(1,4)$, this is $$y-4=2x(1-x)\\ \boxed{y-4=2x-2x^2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.