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Find all values of $x$ at which the tangent line to the given curve has intercept $y = 2$

I am confused about the $y$-intercept $2$

the function $$f(x) = \frac{(2x + 5)}{(x + 2)}$$

The derivative is $$f'(x) = \frac{- 1}{(x^2 + 4x + 4)}$$

Now what should I do next with the given $y$ intercept?

Sorry for a easy question just have misconception on derivative's relationships with a tangent and what is the derivative at a point gives us?

haunted85
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DevX
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3 Answers3

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equation of tangent of f(x):

$$\forall t \in R, y_x(t) =f'(x).t+b$$

Here $b=2$ because it is y-intercept 2 (go through (y,t)=(2,0)):

$$y_x(0) = f'(x).0 + b = 2$$

Finally you have to solve (go through f(x)):

$$y_x(x)=f(x)$$

or again:

$$ f'(x).x+2= f(x)$$

more explicit, you have to solve:

$$2-\frac{x}{x^2 + 4x + 4}=\frac{2x + 5}{x + 2}$$

Katsu
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The general equation of a line passing through ($x_1,y_1$) and having slope m is$$y-y_1=m(x-x_1)$$

The general equation of a tangent to the curve at ($x_1,y_1$) is, $$Y-y_1=- \frac{1}{x^2 + 4x + 4}(X-x_1)$$where $(X,Y)$ is any point on the tangent.

The tangent line passes through (0,2). Hence substituting $X=0$ and $Y=2$ into the above equation, we get $$y_1-2=\frac{1}{4}x_1$$ Now the slope of the tangent is $\frac {1}{4}$. Hence $$\frac {1}{x^2 + 4x + 4}=\frac {1}{4}$$ Solve for x to get the required values.

GTX OC
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    is $x_1=0$ or $x=0$? – Vikram Oct 26 '13 at 14:46
  • Yeah fixed it. Thanks. – GTX OC Oct 26 '13 at 15:58
  • @GTX OC: $2-y_1=\frac{1}{4}x_1$ instead of $y_1-2=\frac{1}{4}x_1$ – Katsu Oct 26 '13 at 18:46
  • @GTOX OC: I don't know how do you go from $y−y1=m(x−x1)$ to $y-y_1=- \frac{1}{x^2 + 4x + 4}(x-x_1)$? I would say $$\forall t \in R, y-y_1=- \frac{1}{t^2 + 4t + 4}(x-x_1)$$ . Because here if the variable of the tangent equation is x then your equation of tangent is not linear. So your solution is wrong. – Katsu Oct 26 '13 at 18:50
  • In $Y-y_1=- \frac{1}{x^2 + 4x + 4}(X-x_1)$, the $x$'s should be $x_1$. – 2'5 9'2 Jul 24 '14 at 08:38
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At $\left(x_1, \dfrac{2x_1+5}{x_1+2}\right)$, the slope of the curve $y=\dfrac{2x+5}{x+2}$ is $\dfrac{-1}{(x_1+2)^2}$. So the equation of a line passing through that point with that slope (in point-slope form) is $$y-\frac{2x_1+5}{x_1+2}=\frac{-1}{(x_1+2)^2}\left(x-x_1\right)$$

If this line has $y$-intercept $2$, then it passes through $(0,2)$. Use that, and then see what $x_1$ has to be. There is only one solution for $x_1$, and hence only one point meeting the required condition.

2'5 9'2
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  • I would be highly thankful if could please tell how having the $y$ intercept as $2$ means that the tangent line has $x$ intercept as $0$. – jiten Jun 22 '19 at 19:22
  • @jiten There is no mention of "$x$ intercept" here. The $y$-intercept is where the curve crosses the $y$-axis. Any place on the $y$-axis has $x$-coordinate $0$. Note that $x$-coordinate of the $y$-intercept is not the same as $x$-intercept. – 2'5 9'2 Jun 22 '19 at 20:22