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I am practicing for a math contest and I encountered the following problem that I don't know how to solve:

For which value of $b$ is there only one intersection between the line $y = x + b$ and the parabola $y = x^2 + 5x + 3$?

How do I solve it?

Edit: I made a mistake typing the problem. It should be $y = x^2 - 5x + 3$ instead of $y = x^2 + 5x + 3 $ the answer key says the answer is -6.

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    $\begingroup$ Do you know how to find a derivative? $\endgroup$ – R. Burton Apr 25 at 0:13
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    $\begingroup$ Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant. $\endgroup$ – the_fox Apr 25 at 0:14
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    $\begingroup$ @NoChance Your observation would be relevant if the line’s equation were of the form $y=b$. The line in the problem is not horizontal, though. $\endgroup$ – amd Apr 25 at 1:37
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    $\begingroup$ @NoChance If you’re going to to that route, then you want the one that has the same slope as the given line. $\endgroup$ – amd Apr 25 at 3:32
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    $\begingroup$ If this problem is for a math contest, I think it's more likely the problem was written with a non-calculus solution in mind. (Also, the question is tagged algebra-precalculus.) $\endgroup$ – YawarRaza7349 Apr 25 at 3:55
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Hint: $$x^2+5x+3=x+b\iff x^2+4x+3-b=0,$$ this quadratic equation should have exactly one solution therefore the discriminant $\Delta=?$

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Note that you like the line $y=x+b$ to be tangent to the parabola. The slope of this line is $ m=1.$ Thus the derivative of your parabola should be the same as the slope of the tangent line. Find the point of the tangency and find the $b$ value so that the line passes through that point.

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The tangent to parabola has slope $y' = 2x+5$ which should be slope of line =1 at the point of contact. So point of contact is at x=-2. Substituting in parabola,we get y=-3. Now this point lies on line as well.

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