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October 2nd, 2019, 04:21 AM   #11
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Math Focus: Area of Circle
But I think answer would be $6x+9$ if question is posted correctly.
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October 2nd, 2019, 06:56 AM   #12
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Quote:
Originally Posted by tahirimanov19 View Post
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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Quote:
Originally Posted by tahirimanov19 View Post
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.
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Last edited by ISP; October 2nd, 2019 at 01:02 PM.
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October 5th, 2019, 10:01 AM   #14
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$\displaystyle Y=X^2$

$\displaystyle dy/dx=2x$ @ 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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