Question about calculas (limit)?

Question

Show there is no tangent to y=x^2 - 4x which passes (2,1)

I used the second fomula to calculate the slope of tangent. the slope of tangent=8. and i got y=8x-15 for the equation of tangent, but how do I show there is no tangent to y=x^2 - 4x??

opps..sry that i've made a mistake....the equation of tangent is supposed to be (y-1)=0(x-2), which y is =1...isnt it?

Answer 1

Slope of tangent =8? Why are you calculating the tangent at x=6? In any case, you are somewhat confused -- there is indeed a tangent line to x²-4x, in fact there is one for every possible value of x (x²-4x is everywhere differentiable). What you want to show is not that there is no tangent, but rather that there is no tangent that passes through (2, 1).

In order to do this, first we note that the general equation for the tangent line passing through x₀ is:

y-f(x₀) = f'(x₀) (x-x₀). This is because the slope of the tangent line at x₀ is f'(x₀), and the line must pass through the point (x₀, f(x₀)), so this is the point-slope form of a line passing through that point with a slope of f'(x₀).

Now, let me write t=x₀, as it is getting tedious to write that subscript. So this equation becomes:

y-f(t) = f'(t) (x-t)

Now, to prove there is no tangent line passing through (2, 1), we suppose the opposite -- that there is such a line, and so for some real value of t, (2, 1) will satisfy the above equation. This means that for some real t:

1-f(t) = f'(t)(2-t)

Now note that f(t) = t²-4t and f'(t) = 2t-4. So substituting this into the equation:

1-(t²-4t) = (2t-4)(2-t)

Now we solve for t:

1-t²+4t = 4t - 2t² - 8 + 4t
-t²+4t+1 = -2t² + 8t - 8
t² - 4t + 9 = 0
t² - 4t + 4 = -5
(t-2)² = -5

And this equation cannot possibly be satisfied by any real value of t. If you wanted to, you could go on and prove that the only complex values of t that satisfy this are 2±i√5, but we don't care, because those aren't real numbers. Thus, there is no tangent to x²-4x which passes through (2, 1).

Answer 2

OK, a tangent line touches the graph (or in different circumstances a shape) at exactly one point, and has the same "slope" at that point. You find the slope of a point by taking the derivative of the equation (which in this case becomes 2x - 4), and plugging in a value for x. For example, the point (0,0) has "slope" -4. This point also has only one tangent line, y= -4x.

To be honest, I don't think any of that matters. Draw a graph of your equation. Note that (2,1) is "inside" the graph. That means any line through it will touch the the graph at 2 points, which is impossible for a tangent line. Except for the verticle line, which isn't an equation.

If that isn't good enough, realize that any line's slope will be rise/run, or if the line is through (2,1) and a point on the graph... (1-x^2+4x)/(2-x)... set this equal to 2x -4 and show that no such x exists. I think that will work, but I haven't checked the math myself.

Answer 3

y = x^2 - 4x
y = x^2 - 4x + 4 - 4
y = (x - 2)^2 - 4
which is a parabola with vertex (2,-4), concave upwards. The point (2,1) lies within the parabola, and all tangents to a parabola necessarily lie without. Therefore, no tangent of y = x^2 - 4x can pass through (2,1).

Another approach is that all tangents of
y = x^2 - 4x are lines such that
(y1 - x^2 + 4x) = (2x - 4)(x1 - x)
y1 - x^2 + 4x = 2x(x1 - x) - 4(x1 - x)
y1 - x^2 + 4x = 2xx1 - 2x^2 - 4x1 - 4x
y1 = 2x1(x - 2) - x^2 - 8x
plugging in y1 = 1 and x1 = 2,
1 = 4x - 8 - x^2 - 8x
x^2 + 4x + 7 = 0
There is no real value of x which will satisfy this equation, so there can be no tangent which passes through (2,1)

Answer 4

if y=8x-15 is tangent, intersect y=x^2-4x in one point
so 8x-15=x^2-4x=>x^2-12x+15=0
y=8x-15 is secant because it have 2 solution

Answer 5

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