# Equation of Tangent Line: Finding at a Given Point | Math HW Help

• teneleven
In summary, the homework equation y = 1 + 2x - x^3 can be solved using either the method in the original post or using Pascal's method. When x = 1, the equation becomes y = x + 1.

## Homework Statement

Find an equation of the tangent line to the curve at the given point.

$$y = 1 + 2x - x^3$$, (1,2)

## Homework Equations

$$y = mx + b$$

$$m = \lim_{x\rightarrow a} \frac{f(x) - f(a)}{x - a}$$

## The Attempt at a Solution

$$\lim_{x\rightarrow 1} \frac{(1 + 2x - x^3) - 2}{x - 1}$$

$$= \lim_{x\rightarrow 1} \frac{-1 + 2x - x^3}{x - 1}$$I'm not sure where to go from there.

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your on the right track, try factoring the top line, the x-1 will cancel out

Rewrite the limit as $$\lim_{x\to 1} -\frac{x^3-2x+1}{x-1}$$ then do some polynomial division.

Thanks for the replies.

Factoring gets me...

$$\lim_{x\rightarrow 1} -x^2 + 1$$

Which results in "0"
Maybe I missed something. :/Using long division I get...

$$m = \lim_{x\rightarrow 1} x^2 + 2x - 1 = 1$$

$$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = 1$$

The point given is (1,2) so...

$$y - 2 = 1(x - 1)$$

$$y = x + 1$$The answer in the book is y = -x + 3

I'm brushing up on my math in preparation for Calc III.

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Try the long division again, you should have gotten it down to
$$m=\lim_{x\to 1} -(x^2+x-1) = -1$$
That leads to the answer in the book.

You could also, since this is "precalculus", use Pascal's method: If y= ax+ b is tangent to 1+ 2x- x3 at x= 1, then ax+ b- (1+ 2x- x3)= x3 +(a-2)x+ (b-1) has a double zero at x= 1. (Since the two graphs cross, clearly x= 1 makes that difference 0, since they are tangent there, it is a double 0.)

Then, setting x= 1, we have 1+ (a-2)+ (b-1)= a+ b- 2= 0 or b= 2- a. We can rewrite the equation as x3+ (a-2)x+ (1-a) and divide by x- 1 to get x2+ x+ (a-1) as quotient. The fact that x= 1 is a double zero of the first polynomial means it must make this 0 also. Taking x= 1 again, 1+ 1+ a-1= a+1= 0 so a= -1. b= 2-a= 2-(-1)= 3. The tangent line is y= -x+ 3.

Of course, it is far easier to note that the derivative of 1+ 2x- x3 is 2- 3x2 and at x= 1, that is -1 so the tangent line is -x+ b and when x= 1, -1+ b= 2 so b= 3. But this is "precalculus"!
(You still have a ways to go to get to Calculus III.)

Calculus 3?! 2 courses of Calculus before you get to the derivative of polynomials? Your joking right?

teneleven did say, in his second post, "I'm brushing up on my math in preparation for Calc III."

Since this was posted in the Precalculus section, and he is taking the derivative by using the basic definition, I interpreted that to me that he is going back and reviewing all of the basics. That's why I said "You still have a ways to go".

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Another relevant equation to use might be:

$$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
Of course, it involves one more step to find f'(1)

Great, thanks for your help. I solved it using the method in my original post.Now I'm trying to solve it using the equation drpizza pointed out.

$$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$After plugging in $$y = -x^3 + 2x + 1$$ into the above equation I get...

$$\lim_{h\to 0} \frac{-(h^3 + 3h^2x + 3hx^2 +2x + 1)}{h}$$

After long division I'm left with...

$$\lim_{h\to 0} -(h^2 + 3x^2 +3hx) - \frac{2x + 1}{h}$$

I'm not sure what to do after this step. The point on the line I'm finding a tangent to is (1,2).

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teneleven said:
Great, thanks for your help. I solved it using the method in my original post.

Now I'm trying to solve it using the equation drpizza pointed out.

$$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$

After plugging in $$y = -x^3 + 2x + 1$$ into the above equation I get...

$$\lim_{h\to 0} \frac{-(h^3 + 3h^2x + 3hx^2 +2x + 1)}{h}$$
No. f(x+h)= -(x+h)3+ 2(x+ h)+ 1= -x3-3hx2-3h2x- 1+ 2x+ 2h+ 1 so f(x+h)- f(x) is
-3hx2-3hx2+ h3+ 2h. All terms that do not involve h cancel out. (do you see why that must be true?) Then (f(x+h)- f(x))/h= -3x2- 3x2+ 2. What is the limit of that as h goes to 0?

After long division I'm left with...

$$\lim_{h\to 0} -(h^2 + 3x^2 +3hx) - \frac{2x + 1}{h}$$

I'm not sure what to do after this step. The point on the line I'm finding a tangent to is (1,2).

## What is the equation of a tangent line?

The equation of a tangent line is a mathematical expression that represents the slope of a line that touches a curve at a single point. It is used to calculate the instantaneous rate of change 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 and the coordinates of the point where it touches the curve. The slope can be calculated using the derivative of the curve at that point. Then, using the point-slope form of a line, the equation of the tangent line can be written as y - y0 = m(x - x0), where (x0, y0) is the point of tangency and m is the slope of the tangent line.

## What is the point-slope form of a line?

The point-slope form of a line is an equation that represents a line using the coordinates of a point on the line and its slope. It is written as y - y0 = m(x - x0), where (x0, y0) is a point on the line and m is the slope of the line.

## What is a derivative?

A derivative is a mathematical concept that represents the instantaneous rate of change of a function at a specific point. It is calculated as the slope of a tangent line to the function at that point. Derivatives are used in many fields of science and engineering, such as physics, economics, and statistics.

## Why do we use the equation of a tangent line?

The equation of a tangent line is used to find the instantaneous rate of change of a curve at a single point. This can be applied in real-world situations where we need to know the exact rate of change at a specific point, rather than an average rate of change over an interval. The equation of a tangent line also helps us to understand the behavior of a curve and make predictions about its future behavior.