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?

michaelcai Jul 17, 2017

#1**0 **

I would assume that only a straight verticle line can intercept with a parabola.

DeadRight Jul 17, 2017

#2**+2 **

Best Answer

The straight line must be tangent to the parabola.

Equate the two functions:

\(x^2+2x+7=6x+b\\x^2-4x+7-b=0\\\)

The discriminant must be zero if there is to be a single solution, so

\((-4)^2-4(7-b)=0\\-12+4b=0\\b=3\)

Alan Jul 17, 2017

#3**+2 **

Here's another approach

If a line intersects a parabola at only one point, it is tangent to the parabola at that point.

So....the slopes are equal at that point

The slope of the parabola at any point is 2x + 2

The slope of the line is a constant, 6

Equate the slopes

2x + 2 = 6

2x = 4

x = 2

So.....this is the value where the slopes are the same.....subbing this into both functions gives us

(2)^2 + 2(2) + 7 = 6(2) + b

15 = 12 + b

b = 3

CPhill Jul 17, 2017

#5**+1 **

Remember that in the Quadratic Formula the discriminant (the part under the square root ), b^2 - 4ac, gives us some info about the solutions we can expect

If the discriminant evaluates to 0, it means that we have a "double root", i.e., only one solution

And since we only * want* one solution point in this problem - the tangent point of the line to the parabola - we can see what value of "b" gives us a discriminant of 0

So... using x^2 - 4x + 7 - b = 0 let's change "b" to "m" so that we don't get it confused wih the "b" in the quadratic formula

So we have x^2 - 4x + 7 - m = 0

And in the quadratic formula, let a = 1, b = -4 and c = 7 - m

So.... the discriminant becomes b^2 - 4ac → (-4)^2 - 4(1)(7 - m)

Now....set this to 0 and solve for "m"

(-4)^2 - 4(1)(7 - m) = 0 simplify

16- 28 + 4m = 0

-12 + 4m = 0

4m = 12

m = 3

So.....the "m" - or in this case, the "b" - that gives us a single solution is 3

CPhill Jul 18, 2017