Find the equations of all the the tangents to the graph of the function f(x) =x^2 -4x +25 that pass thru the orgin (0,0).

I'm not sure what to do here. I know f'(x) = 2x - 4. But I don't know what to do with that knowledge.

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- Jun 12th 2006, 08:00 PM #1

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- Jun 12th 2006, 08:16 PM #2

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Originally Posted by**becky**

$\displaystyle y=mx$,

such a line meets the parabola:

$\displaystyle y=x^2 -4x +25$

when:

$\displaystyle mx=x^2 -4x +25$,

and if the line is a tangent this last equation has a single root

(since in general a line cuts a parabola at two points, one point

or no points, the one point case is a tangent).

Rearranging the equation gives:

$\displaystyle x^2-(m+4)x+25=0$,

and this has a single root when its discriminant is zero:

$\displaystyle (m+4)^2-4\times 25=0$,

this is because a general quadratic equation $\displaystyle ax^2+bx+c=0$ has

roots $\displaystyle x=(-b\pm \sqrt{b^2-4a.c})/(2a)$, and both

of the roots are coincident (ie there is only one root) when

$\displaystyle b^2-4a.c=0$.

Solve this for $\displaystyle m$ and then plug these solutions back into the

equation of a line through the origin to get the equations for all

the the tangents to the parabola which pass through the origin.

RonL

- Jun 12th 2006, 11:36 PM #3

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Originally Posted by**becky**

f(x) = x^2 -4x +25

So the f'(x) is 2x -4.

What is f'(x)? In relation to all tangents to the graph of f(x)?

f'(x) is the slope of all those tangent lines.

So for the tangent lines that pass through the origin (0,0), you have a known point, and the the slope of the tangent lines.

Using the point-slope form of the equation of a line,

(y -y1) = m(x -x1) -------**

with (0,0) as (x1,y1), and (2x -4) as m,

(y -0) = (2x -4)(x -0)

y = (2x -4)*x ---------(i), the equation of all tangent lines that pass through (0,0).

We did not expand the righthand side into (2x^2 -4x) because that will be confusing. A tangent line is a straight line, and y = 2x^2 -4x is not a straight line

Each of those tangent lines have a point in common with f(x). That is the point of tangency. At that point, the coordinates of f(x) and of the tangent line are one and the same. So, the y-coordinate of f(x) is the same as the y-coordinate of the tangent line.

The y of f(x) is f(x). It is (x^2 -4x +25).

The y of the tangent line is (2x-4)x

Since those two y's are the same, then,

(2x -4)x = x^2 -4x +25

Simplifying,

2x^2 -4x = x^2 -4x +25

x^2 = 25

x = +,-5

when x = 5,

m = f'(5) = 2(5) -4 = 6

So, the tangent line is

y = (6)x

y = 6x ---------------------answer.

when x = -5,

m = f'(-5) = 2(-5) -4 = -14

So, the tangent line is

y = (-14)x

y = -14x ---------------------answer.

Therefore, there are only two of those tangent lines to f(x) = x^2 -4x +25 that pass through the origin (0,0).