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 October 2nd, 2019, 04:21 AM #11 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle But I think answer would be $6x+9$ if question is posted correctly.
October 2nd, 2019, 06:56 AM   #12
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Quote:
 Originally Posted by tahirimanov19 The function of slop would be $g(x)=ax+b$. Let the tangent point be $(x_1, y_1)$. Therefore, $y_1=ax_1+b={x_1}^2$. $y'=2x$ therefore, slope of g(x) is $2x_1$. $2{x_1}^2+b={x_1}^2 \Rightarrow b=-{x_1}^2$. $g(x)=2{x_1}x-{x_1}^2$ $g(3)=6{x_1}-{x_1}^2=4 \Rightarrow x_1=3 \pm \sqrt{5}$ $g(x)=(3 \pm \sqrt{5})x - (3 \pm \sqrt{5})^2$ Which means there is no tangent that passes through $(3,4)$.

$y - 4 = (6+2\sqrt{5})(x-3)$

$y - 4 = (6-2\sqrt{5})(x-3)$
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October 2nd, 2019, 12:59 PM   #13
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From: SV USA

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Quote:
 Originally Posted by tahirimanov19 Which means there is no tangent that passes through $(3,4)$.
Wouldn't common sense lead you to realize there must be at least one line that is tangent to $y = x^2$ and that passes through $(3,4)$? Am I missing something here...?

Edit: I see Skeeter beat me to it above. Thank you.

Last edited by ISP; October 2nd, 2019 at 01:02 PM.

 October 5th, 2019, 10:01 AM #14 Newbie   Joined: Oct 2019 From: India Posts: 2 Thanks: 2 $\displaystyle Y=X^2$ $\displaystyle dy/dx=2x$ @ P(3,4) The value of $\displaystyle dy/dx = 6$ now using the slope tangent equation of line : $\displaystyle (Y-9)=6(X-3)$ $\displaystyle Y-9=6X -18$ $\displaystyle Y=6X-9$ Last edited by jinxrifle; October 5th, 2019 at 10:06 AM.

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