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For what value of b will the line y=-2x+b be tangent to the parabola y=3x^2+4x-1
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There are TWO WAYS to solve the problem: one way is Algebra, the other way is Calculus.
I will show you both.
1. Calculus way
The slope of the given line is -2, the constant value.
The slope of the parabola is 6x+4 at the point with abscissa x.
In order for the line be tangent to the parabola, the necessary condition is "slope" = "slope",
which gives -2 = 6x + 4, and then x= -1.
At x= -1, the y-coordinate of the parabola is 3*(-1)^2 + 4*(-1) - 1 = 3*1 - 4 - 1 = -2;
the y-coordinate of the line is -2*(-1) + b= 2 + b.
Y-coordinate should be the same, which gives 2 + b = -2, and hence b= -4. ANSWER
So, the Calculus solution is completed.
2. Algebra way
You write this equation
-2x + b = 3x^2 + 4x-1
to find common point of the line and the parabola.
Its standard form is
3x^2 + 6x - (1+b) = 0.
The straight line is a tangent to the parabola if and only if there is ONLY ONE common point;
in other words, if there is ONLY ONE solution to the last equation.
It means that the discriminant of this equation is zero
6^2 + 4*3*(1+b) = 0,
or
36 + 12*(1+b) = 0
3 + (1+b) = 0
b = -4.
We got the same answer.
Thus the problem is solved and two basic solutions are presented.