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September 30th, 2019, 07:47 PM   #1
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Find the slope of the parabola y=x^2 at the point P(3,4), and use it to find the equation of the tangent line to y=x^ at x=3.
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September 30th, 2019, 08:23 PM   #2
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Ok I did it. Now tell me your answer so I can check my work.
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September 30th, 2019, 09:45 PM   #3
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Find the slope of the parabola y=x^2 at the point P(3,4), and use it to find the equation of the tangent line to y=x^ at x=3.
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Ok I did it. Now tell me your answer so I can check my work.
Well I couldn't do it! (3, 4) isn't even a point on the parabola!

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October 1st, 2019, 04:27 PM   #4
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Well I couldn't do it! (3, 4) isn't even a point on the parabola!

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lol that's what I get for being snarky.
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October 1st, 2019, 07:22 PM   #5
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Hmmmm... I wonder if there's a metric that will do the trick so that the point really is on the parabola. Have to think that one over.

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October 1st, 2019, 11:44 PM   #6
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it is probably P(3,9).
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October 2nd, 2019, 02:40 AM   #7
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it is probably P(3,9).
Nah! That would be too easy.

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October 2nd, 2019, 02:43 AM   #8
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Find the slope of the parabola y=x^2 at the point P(3,4), and use it to find the equation of the tangent line to y=x^ at x=3.
Okay, back to the actual question.

The slope of the line tangent to the curve at (3, 9) (or whatever) is the value of the first derivative at that point. So $\displaystyle y = x^2 \implies \dfrac{dy}{dx} = 2x$.

That gives you the slope of a line (the derivative) and a point on that line (3, 9). How can you find the equation of the line from here?

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October 2nd, 2019, 03:00 AM   #9
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Perhaps the coordinates were correct but the OP mistakenly thought it was on the curve. Then the question is "find the equation of the tangent line to $\displaystyle y=x^2$ that goes through $\displaystyle (3,4)$."

If that was the question then I get two solutions: $\displaystyle y=(6 \pm 2\sqrt{5})(x-3)+4$
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October 2nd, 2019, 04:20 AM   #10
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find the equation of the tangent line to $\displaystyle y=x^2$ that goes through $\displaystyle (3,4)$
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 $(3,4)$.
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