Calculus tangent line problem.

In summary, to find the equations of all tangents to the graph f(x) = x^2-4x + 25 that pass through the origin, one must first find the derivative using the slope formula and the given point (0,0). After simplifying, the equation for slope is set equal to the derivative to find the values of x where the tangent line will pass through (0,0). The final equation for the tangent lines is y=(2a-4)x, where a is any point on the graph.
  • #1
CJSGrailKnigh
57
0
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.
 
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  • #2
One small issue the derivative should be 2x-4 should it not?
 
  • #3
ya i just copied my work wrong sorry.
 
  • #4
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.
 
  • #5
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.
 

Related to Calculus tangent line problem.

What is a tangent line in calculus?

A tangent line in calculus is a line that touches a curve at a single point, and has the same slope as the curve at that point. It can be used to approximate the behavior of a curve at a specific point.

How do you find the equation of a tangent line?

To find the equation of a tangent line, you need to know the slope of the tangent line at the point of interest. This can be found by taking the derivative of the function at that point. Then, you can use the point-slope form of a line to write the equation of the tangent line.

What is the point of finding a tangent line in calculus?

Finding a tangent line in calculus can be useful for approximating the behavior of a curve at a specific point. It can also be used to find the slope of a curve at a given point, which can provide valuable information about the behavior of the curve.

How is a tangent line different from a secant line?

A tangent line touches a curve at a single point, while a secant line intersects a curve at two points. The slope of a tangent line is the slope of the curve at the point of tangency, while the slope of a secant line is the average rate of change between the two points of intersection.

What is the relationship between a tangent line and the derivative of a function?

The derivative of a function at a point is equal to the slope of the tangent line at that point. This means that the derivative can be used to find the slope of the tangent line and the equation of the tangent line at a specific point on the curve.

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