finding a tangent line to a parabola

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.

• Do you know how to find a derivative? – R. Burton Apr 25 at 0:13
• Simply $x^2+5x+3=x+b$ must have only one solution, i.e. $x^2+4x+(3-b)=0$ should have zero discriminant. – the_fox Apr 25 at 0:14
• @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. – amd Apr 25 at 1:37
• @NoChance If you’re going to to that route, then you want the one that has the same slope as the given line. – amd Apr 25 at 3:32
• 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.) – YawarRaza7349 Apr 25 at 3:55

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 $$\Delta=?$$
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.
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.