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.


$(2+h)^2-1^2/h$ 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

  • 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

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.

  • $\begingroup$ 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

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 $$y-4=2x(1-x)\\ \boxed{y-4=2x-2x^2}$$


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