Finding a tangent in a Cubic Function

In summary, to find the tangent at the point x=2.5 for the cubic function y=x^3-16x^2+69x-54, you first need to find the derivative, which is 3x^2-32x+69. Then, plug in x=2.5 to get the slope of the tangent, which is 34.125. Using the point-slope formula, you can then find the equation of the tangent line.
  • #1
jahaddow
47
0

Homework Statement


My cubic function is y=(x-6)(x-1)(x-9) or y=x^3-16x^2+69x-54
I need to find the tangent at the point x=2.5

Homework Equations


The Attempt at a Solution


All that I have managed to do is work out the y value for x=2.5, that is y=34.125
Please help someone!
 
Last edited:
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  • #2
the gradient is the derivative of your function, so differentiate, ie. find [itex]\frac{dy}{dx}[/itex]
same as other post
 
  • #3
Derive and plug in x = 2.5

In other words, solve for [tex]\frac{dy}{dx}[/tex], and plug in 2.5 for x:

y = x3-16x2+69x-54

[tex]\frac{dy}{dx}[/tex] = 3x2-32x+69

At x = 2.5, [tex]\frac{dy}{dx}[/tex], which is the tangent, equals

3(2.5)2-32(2.5)+69 = ?

I am too lazy to do the calculation, but here is the basic setup.

If you are looking for a tangent line, then use y'(2.5) as the slope for the slope

equation: y-y0 = f'(2.5)(x-x0)
 
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  • #4
thankyou ever so much, if I have any more questions I will ask!
 
  • #5
Ok, so I now have the Gradient and the X and Y values of the tangent. How do I get the equation of the tangent.

ps. Realy sorry for all the trouble, you are a big help!
 
  • #6
Well, if you have the slope of a line and a point on the line, how do you find the equation of the line? Remember back to algebra!
 
  • #7
i can't think back that far, Please explain
 
  • #8
ahh I know now, Point slope formula!
 

1. How do you find the tangent of a cubic function?

To find the tangent of a cubic function, you can use the process of differentiation. This involves finding the derivative of the function, which will give you the slope of the tangent line at any given point. The equation for the tangent line can then be found using the point-slope formula.

2. What is the difference between a tangent and a secant in a cubic function?

A tangent is a line that touches the curve of a cubic function at one point, while a secant is a line that intersects the curve at two points. A tangent line has a slope that is equal to the derivative of the function at the point of tangency, while a secant line has an average slope between the two points of intersection.

3. Can a cubic function have more than one tangent line?

Yes, a cubic function can have more than one tangent line. This is because a cubic function can have multiple points of inflection where the curvature changes, resulting in multiple tangent lines with different slopes.

4. How do you find the point of tangency on a cubic function?

To find the point of tangency on a cubic function, you can set the derivative of the function equal to the slope of the tangent line. This will give you an equation to solve for the x-value of the point of tangency. You can then plug this x-value into the original function to find the y-value.

5. Can a cubic function have a horizontal tangent line?

Yes, a cubic function can have a horizontal tangent line. This occurs when the derivative of the function is equal to 0, resulting in a slope of 0 for the tangent line. This can happen at points of inflection or at the maximum or minimum points of the function.

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