- #1

CJSGrailKnigh

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**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

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.

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.

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