Calculus tangent line problem.

[CJSGrailKnigh]

1. Find the equations of all tangents to the graph f(x) = x^2-4x + 25 that pass through the origin.
2. The relevant equations to my knowledge are the slope formula and f ' (x)
3. First I found the derivative using various rules to be f ' (x) = 2x - 4
after this i used the slope formula and my given point (0,0) to fill in this equation for slope
m = 0 - X^2 +4x - 25
0-x

I then simplified to this by canceling the negatives of the numerator and denominator and by then moving the denominator to the top by making the exponent a negative and then multiplied the trinomial by the monomial to get m = x +4 - (25/x)

after this I set m = f ' (x) in order to find the values of x where the tangent line will pass through (0,0).

I got -x-(25/x) = 0 however this function does not have any zeros and therefore can not be the equation I'm looking for because the slopes i should get according to the answer key are m = -14 and m = 6

Any help would be greatly appreciated.

 

[GNelson]

One small issue the derivative should be 2x-4 should it not?

 

[CJSGrailKnigh]

ya i just copied my work wrong sorry.

 

[CJSGrailKnigh]

The Answer is obtained in the way I had started however the two mistakes I made were first I should have done delta y as being : (x^2-4x+25) -0 and the delta x being x-0 then equating the derivative to the slop formula to get:

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

then cross multiplying to get:

x^2-4x+25 = 2x^2 - 4x or x^2-25 = 0

which is factored by difference of squares to be (x-5)(x+5) = 0 and therefore when x = 5 or -5 the line of tangency will pass through the origin and one can find the equations for the lines.

 

[sutupidmath]

since m=f'(a)=2a-4, you could use the slope-point formula for a line, that is:

y-y_1=m(x-x_1) since the tangent passes through the origin that is one point on the tangent line is (0,0) so
y-0=f'(a)(x-0), from here

y=(2a-4)x, where a is any point on the graph.