Quadratic Equation Help

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

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

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

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

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