SOLUTION: If the parabola y_1 = x^2 + 2x + 7 and the line y_2 = 6x + b intersect at only one point, what is the value of b?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: If the parabola y_1 = x^2 + 2x + 7 and the line y_2 = 6x + b intersect at only one point, what is the value of b?      Log On


   



Question 1086039: If the parabola y_1 = x^2 + 2x + 7 and the line y_2 = 6x + b intersect at only one point, what is the value of b?
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
If they only intersect at one point then the line is tangent to the parabola at the intersection point.
Find the derivative of the parabola, the value of the derivative is equal to the slope of the tangent line at that point.
Find all values where
dy%2Fdx=m=6
For the parabola,
dy%2Fdx=2x%2B2=6
2x=4
x=2
So then find the value of the parabola when x=2,
y=x%5E2%2B2x%2B7
y=%282%29%5E2%2B2%282%29%2B7
y=4%2B4%2B7
y=15
The line also intersects the parabola at this point, so then,
6%282%29%2Bb=15
b=3
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Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
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This problem is for Algebra-I, not for Calculus.
So I will give here another (more adequate) solution.

To get the common points, substitute y = 6x + b into the first equation.  You will get

6x + b = x%5E2+%2B+2x+%2B+7.


It is the same as

x%5E2+-+4x+%2B+7-b = 0.   (1)


The original system has a unique solution if and only if the equation (1) has a unique solution.

It happens if and only if the discriminant of the equation (1) is zero:

d = 4%5E2+-+4%2A%287-b%29 = 0,   or

16 = 4*(7-b),   or  4 = 7-b,   or   b = 7-4 = 3.


Answer.  b = 3.

Solved.