Is Y=23x+18 the Correct Equation for the Tangent Line at X=2?

In summary, the conversation is about finding the equation of the tangent line at x=2 for the given function. The process involves finding the derivative, which is done incorrectly at first, but is corrected. The final answer is determined to be y=23x+18, which can be confirmed by plugging in values and solving for the y-intercept.
  • #1
chjopl
21
0
Find the equation of the tangent line at x=2
F(X)=2x+7x(e^(x-2))

I know you put 2 into find the y intercept and you take the derivative to find the slope. I get 2+(7e^(-2)x + 7e^-2)e^2 for my derivative and y=23x+18 as my answer and i know that is not correct.
 
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  • #2
You derivative looks off. Remember,

[tex]\frac{d}{dx}e^{x-2}=e^{x-2}\frac{d}{dx}(x-2)=e^{x-2}(1)=e^{x-2}[/tex]

and apply the product rule carefully!


edit-ahh, Galileo is right, your slope is fine. Your posted derivative may be a casualty of ascii.
 
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  • #3
You got the slope right. So y=23x+b. Now use the fact that the line passes through the point (2,f(2)) to find b.
 
  • #4
chjopl said:
Find the equation of the tangent line at x=2
F(X)=2x+7x(e^(x-2))

I know you put 2 into find the y intercept and you take the derivative to find the slope. I get 2+(7e^(-2)x + 7e^-2)e^2 for my derivative and y=23x+18 as my answer and i know that is not correct.

The derivative should be:
[tex] f'(x)=2+[7\exp({-2})](\exp{x}+x\exp{x) [/tex]
Insert 2 to find the slope.It's 23.
Compute f(2).It's 18.With the last 2 numbers,u should be able to find the equation for the tangent line.

We're just like eagles when it comes to simple problems.Three almost identical posts at the same time...
 
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  • #5
so was y=23x+18 correct?
 
  • #6
chjopl said:
so was y=23x+18 correct?

Obviously not,try to follow my post correctly.It's easy maths,after all.
 
  • #7
chjopl said:
so was y=23x+18 correct?

What is 23(2)+ 18?
 

FAQ: Is Y=23x+18 the Correct Equation for the Tangent Line at X=2?

What is the equation of tangent line?

The equation of tangent line is a mathematical expression that represents the slope of a curve at a specific point. It is used to find the instantaneous rate of change of a function at that point.

How do you find the equation of tangent line?

To find the equation of tangent line, you need to find the slope of the curve at the given point. This can be done by taking the derivative of the function at that point. Once you have the slope, you can plug it into the point-slope form of a line to get the equation of the tangent line.

What is the point-slope form of a line?

The point-slope form of a line is y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is any point on the line. This form is useful for finding the equation of tangent line as it allows us to plug in the slope and a known point on the line.

What is the significance of the equation of tangent line?

The equation of tangent line is significant as it helps us understand the behavior of a curve at a specific point. It also allows us to calculate the instantaneous rate of change of a function, which is important in many scientific and engineering applications.

Can the equation of tangent line be used for any type of curve?

Yes, the equation of tangent line can be used for any type of curve, including polynomial, exponential, logarithmic, and trigonometric functions. It is a general method for finding the slope of a curve at a given point.

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