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Find the tangent line on $f(x)=x^2$ which runs parallel to the slope of $y=2x+6$.

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  • $\begingroup$ The slope of y=2x+6 is 2, and y=2 is precisely a horizontal line. When is the tangent line to f(x)=$x^2$ horizontal? $\endgroup$ – Äres Dec 22 '19 at 10:01
  • $\begingroup$ This is not the right equation. $\endgroup$ – James Dec 22 '19 at 10:31
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    $\begingroup$ Welcome to MathSE. When you pose a question here, it is expected you include your own thoughts on the problem. You should show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Dec 22 '19 at 11:16
  • $\begingroup$ See How To Ask A Homework Question. That aside, this question has been asked here many times, albeit with different numbers. A quick search will find several examples. [This answer](math.stackexchange.com/a/484877/265466 ) in particular is quite detailed and complete. You might also look through the related questions in the list at right. $\endgroup$ – amd Dec 22 '19 at 19:06
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    $\begingroup$ Does this answer your question? Tangent Line to the curve that is Parallel to the line $\endgroup$ – amd Dec 22 '19 at 19:07
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Solution 1: The equation of the tangent line is $y= 2x -1$

Explanation: Suppose $y= 2x + c .......(1)$

be the equation of the tangent we're looking for. Why the slope of the line is $2$? Think about it. Since this straight line is a tangent to the graph of $y= x^2$........(2), there is exactly one solution to the equation $(1)$ and $(2)$. If we substitute the $y$ value from (1) into the equation (2), then we have the following equation

$$ x^2 = 2x +c.$$ Equivalently, $$ x^2 -2x -c.$$ This quadratic equation must have discriminant equals to $0$. Think about it why? Therefore, we have $$(-2)^2 - 4.1. (-c)= 0.$$ Hence, $$ c= -1$$. This gives our required tangent line.

Solution 2: Easy one

Let $(p, p^2)$ be point on the graph where the tangent line touches the graph.

Note that the derivative of the function $x^2$ at the point $(p, p^2)$ is $2p$. You see why? This derivative is essentially the slope of the tangent line. Hence $$2p = 2.$$ Therefore, $p =1$. Then the point on the graph is basically $(1,1)$. Now use our favorite point-slope formula to find the equation of the tangent line. So, the equation of the tangent line is $$y-1 = 2(x-2),$$ and we're done.

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