For what value of b will the line y=-2x+b be tangent to the parabola y=3x^2+4x-1? Show work-2 marks
you ever gonna show any of your work?
you want the slope of the line (-2) to be the same as the slope of the parabola.
The slope of the parabola at any point (x,y) is 6x+4
So, if 6x+4 = -2, x = -1
y(-1) = 6, so the line
y = -2x+b must pass through (-1,-2)
I think now you can find b.
Check your work at
http://www.wolframalpha.com/input/?i=plot+y%3D3x^2%2B4x-1%2C+y%3D-2x-4
y(-1) = -2
Well, well, well, let's find out! When a line is tangent to a parabola, that means it only touches the parabola at one point, like a fancy fingertip boop. So let's find the point where they make sweet, sweet contact.
We have the line equation y = -2x + b and the parabola equation y = 3x^2 + 4x - 1. To find the point of contact, we need the x-coordinate to be the same in both equations.
So, let's equate them:
-2x + b = 3x^2 + 4x - 1
Now, to solve this quadratic equation, let's rearrange it to the classic form:
3x^2 + 6x + (4x - 2 - b) = 0
To have only one solution for x, the discriminant (b^2 - 4ac) should be equal to zero. In this case, a = 3, b = 6, and c = (4x - 2 - b).
Let's plug the values into the quadratic formula, which is a bit like a math magician's hat:
b^2 - 4ac = 0
6^2 - 4(3)(4x - 2 - b) = 0
Simplifying things a bit:
36 - 12(4x - 2 - b) = 0
Now, let's do some beastly expanding:
36 - 48x + 24 - 12b = 0
Combining like terms:
60 - 48x - 12b = 0
To be a tangent, the line should touch the parabola at one point only. That means we have one solution for x, and that can only happen when the coefficient of x is zero:
-48x = 0
Solving this equation gives us:
x = 0
Now, let's substitute x = 0 into either the line or parabola equation. I'll choose the line equation to save time:
y = -2(0) + b
y = b
So when x = 0, y = b. That's the point of contact!
Therefore, the line y = -2x + b will be tangent to the parabola y = 3x^2 + 4x - 1 when b = 0. Ha! Zero is quite the star, isn't it?
To find the value of b that will make the line y=-2x+b tangent to the parabola y=3x^2+4x-1, we need to set the equation of the line equal to the equation of the parabola and find the value of x at the point of tangency.
First, we set the two equations equal to each other:
-2x + b = 3x^2 + 4x - 1
Next, we rearrange the equation to form a quadratic equation:
3x^2 + 4x -2x - b + 1 = 0
Simplifying further, we get:
3x^2 + 2x - (b - 1) = 0
To find the value of x at the point of tangency, we know that the quadratic equation must have only one solution. This means that the discriminant (b^2 - 4ac) must be equal to zero.
The discriminant is given by:
b^2 - 4ac = 0
For our equation, the coefficients are:
a = 3
b = 2
c = -(b - 1)
Substituting these values into the discriminant equation:
(2)^2 - 4(3)(-(b - 1)) = 0
4 - 12(-(b - 1)) = 0
Expanding and simplifying:
4 + 12b - 12 = 0
12b - 8 = 0
12b = 8
b = 8/12
Reducing the fraction:
b = 2/3
Therefore, the value of b that will make the line y=-2x+b tangent to the parabola y=3x^2+4x-1 is b = 2/3.