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?
I would assume that only a straight verticle line can intercept with a parabola.
The straight line must be tangent to the parabola.
Equate the two functions:
The discriminant must be zero if there is to be a single solution, so
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