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Determine the equation of the line that is tangent to the parabola with equation

$y = x^2 − 2x + 2$

at the point $(3, 5)$

Henno Brandsma
  • 242,131
  • Joachimsthal notation for $Ax_0x + B(x_0y + xy_0) + Cy_0y + F(x_0 + x) + G(y_0 + y) + H=0$ where $(x_0,y_0)=(3,5)$ works here: $(y+5)/2=3x-2(x+3)/2+2.$ Admittedly not the first choice though... – Jan-Magnus Økland Apr 22 '20 at 18:57
  • Duplicate of https://math.stackexchange.com/q/92165/265466 and many others. – amd Apr 22 '20 at 19:38

3 Answers3

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The directional coefficient equals $\frac{dy}{dx} = 2x-2$ evaluated at $x=3$.

Henno Brandsma
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  1. Find $\frac{dy}{dx}$.

  2. Insert $x=3$ into $\frac{dy}{dx}$. This will give you the gradient which equals $a$ of the tangent equation $ax+c$.

  3. Next, insert $y=5$ into your tangent equation and rearrange to find $c$.

Try to do it yourself but ask questions if you get stuck.

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For fun:

0) $y-5=m(x-3)$; Line passes through $(3,5)$;

1) Intersection of line with $y=x^2-2x+2$;

$m(x-3)+5=x^2-2x+2;$

$x^2-x(2+m)+(3m-3)=0$;

2) Discriminant:

$D= (2+m)^2-4(3m-3)=0$;

$m^2-8m+16=0;$

$(m-4)^2=0$; $m=4$;

3) Tangent:

$y-5=4(x-3).$

Peter Szilas
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