Find the Equation of a Line Tangent to a Curve At a Given Point - Pre-Calculus

We begin by recalling that one way of defining the derivative of a function is the slope of the tangent line of the function at a given point. Therefore, finding the derivative of our equation will allow us to find the slope of the tangent line. Since the two things needed to find the equation of a line are the slope and a point, we would be halfway done.

We calculate the derivative using the power rule.

However, we don't want the slope of the tangent line at just any point but rather specifically at the point . To obtain this, we simply substitute our x-value 1 into the derivative.

Therefore, the slope of our tangent line is .

We now need a point on our tangent line. Our choices are quite limited, as the only point on the tangent line that we know is the point where it intersects our original graph, namely the point .

Therefore, we can plug these coordinates along with our slope into the general point-slope form to find the equation.

Solving for will give us our slope-intercept form.

The equation of the tangent line at depends on the derivative at that point and the function value.

The derivative at that point of is

using the Power Rule

which means

The derivative is zero, so the tangent line will be horizontal.

It intersects it at since , so that line is .

Rewrite in slope-intercept form, , to determine the slope.

The slope of the given function is 2.

Substitute the slope and the given point, , in the slope-intercept form to determine the y-intercept.

Substitute this and the slope back to the slope-intercept equation.

The equation of the tangent line is:

We begin by finding the equation of the derivative using the limit definition:

We define and as follows:

We can then define their difference:

Then, we divide by h to prepare to take the limit:

Then, the limit will give us the equation of the derivative.

Now, we must realize that the slope of the line tangent to the curve at the given point is equivalent to the derivative at the point. So if we define our tangent line as: , then this m is defined thus:

Therefore, the equation of the line tangent to the curve at the given point is:

First, find the slope of this tangent line by taking the derivative:

Plugging in 1 for x:

So the slope is 4

Now we need to find the y-coordinate when x is 1, so plug 1 in to the original equation:

To write the equation, use point-slope form and then use algebra to change to slope-intercept like the answer choices:

distribute the 4

add 2 to both sides

First, find the slope of the tangent line by taking the first derivative:

To finish determining the slope, plug in the x-value, 2:

the slope is 6

Now find the y-coordinate where x is 2 by plugging in 2 to the original equation:

To write the equation, start in point-slope form and then use algebra to get it into slope-intercept like the answer choices.

distribute the 6

add 8 to both sides

First, take the first derivative in order to find the slope:

To continue finding the slope, plug in the x-value, -2:

Then find the y-coordinate by plugging -2 into the original equation:

The y-coordinate is

Now write the equation in point-slope form then algebraically manipulate it to match one of the slope-intercept forms of the answer choices.

distribute the -5

add to both sides

First distribute the . That will make it easier to take the derivative:

Now take the derivative of the equation:

To find the slope, plug in the x-value -3:

To find the y-coordinate of the point, plug in the x-value into the original equation:

Now write the equation in point-slope, then use algebra to get it into slope-intercept like the answer choices:

distribute

subtract from both sides

write as a mixed number

We begin by recalling that one way of defining the derivative of a function is the slope of the tangent line of the function at a given point. Therefore, finding the derivative of our equation will allow us to find the slope of the tangent line. Since the two things needed to find the equation of a line are the slope and a point, we would be halfway done.

We calculate the derivative using the power rule.

However, we don't want the slope of the tangent line at just any point but rather specifically at the point . To obtain this, we simply substitute our x-value 1 into the derivative.

Therefore, the slope of our tangent line is .

We now need a point on our tangent line. Our choices are quite limited, as the only point on the tangent line that we know is the point where it intersects our original graph, namely the point .

Therefore, we can plug these coordinates along with our slope into the general point-slope form to find the equation.

Solving for will give us our slope-intercept form.

The equation of the tangent line at depends on the derivative at that point and the function value.

The derivative at that point of is

using the Power Rule

which means

The derivative is zero, so the tangent line will be horizontal.

It intersects it at since , so that line is .

Rewrite in slope-intercept form, , to determine the slope.

The slope of the given function is 2.

Substitute the slope and the given point, , in the slope-intercept form to determine the y-intercept.

Substitute this and the slope back to the slope-intercept equation.

The equation of the tangent line is:

We begin by finding the equation of the derivative using the limit definition:

We define and as follows:

We can then define their difference:

Then, we divide by h to prepare to take the limit:

Then, the limit will give us the equation of the derivative.

Now, we must realize that the slope of the line tangent to the curve at the given point is equivalent to the derivative at the point. So if we define our tangent line as: , then this m is defined thus:

Therefore, the equation of the line tangent to the curve at the given point is:

First, find the slope of this tangent line by taking the derivative:

Plugging in 1 for x:

So the slope is 4

Now we need to find the y-coordinate when x is 1, so plug 1 in to the original equation:

To write the equation, use point-slope form and then use algebra to change to slope-intercept like the answer choices:

distribute the 4

add 2 to both sides

First, find the slope of the tangent line by taking the first derivative:

To finish determining the slope, plug in the x-value, 2:

the slope is 6

Now find the y-coordinate where x is 2 by plugging in 2 to the original equation:

To write the equation, start in point-slope form and then use algebra to get it into slope-intercept like the answer choices.

distribute the 6

add 8 to both sides

First, take the first derivative in order to find the slope:

To continue finding the slope, plug in the x-value, -2:

Then find the y-coordinate by plugging -2 into the original equation:

The y-coordinate is

Now write the equation in point-slope form then algebraically manipulate it to match one of the slope-intercept forms of the answer choices.

distribute the -5

add to both sides

First distribute the . That will make it easier to take the derivative:

Now take the derivative of the equation:

To find the slope, plug in the x-value -3:

To find the y-coordinate of the point, plug in the x-value into the original equation:

Now write the equation in point-slope, then use algebra to get it into slope-intercept like the answer choices:

distribute

subtract from both sides

write as a mixed number

We begin by recalling that one way of defining the derivative of a function is the slope of the tangent line of the function at a given point. Therefore, finding the derivative of our equation will allow us to find the slope of the tangent line. Since the two things needed to find the equation of a line are the slope and a point, we would be halfway done.

We calculate the derivative using the power rule.

However, we don't want the slope of the tangent line at just any point but rather specifically at the point . To obtain this, we simply substitute our x-value 1 into the derivative.

Therefore, the slope of our tangent line is .

We now need a point on our tangent line. Our choices are quite limited, as the only point on the tangent line that we know is the point where it intersects our original graph, namely the point .

Therefore, we can plug these coordinates along with our slope into the general point-slope form to find the equation.

Solving for will give us our slope-intercept form.

The equation of the tangent line at depends on the derivative at that point and the function value.

The derivative at that point of is

using the Power Rule

which means

The derivative is zero, so the tangent line will be horizontal.

It intersects it at since , so that line is .

Rewrite in slope-intercept form, , to determine the slope.

The slope of the given function is 2.

Substitute the slope and the given point, , in the slope-intercept form to determine the y-intercept.

Substitute this and the slope back to the slope-intercept equation.

The equation of the tangent line is:

We begin by finding the equation of the derivative using the limit definition:

We define and as follows:

We can then define their difference:

Then, we divide by h to prepare to take the limit:

Then, the limit will give us the equation of the derivative.

Now, we must realize that the slope of the line tangent to the curve at the given point is equivalent to the derivative at the point. So if we define our tangent line as: , then this m is defined thus:

Therefore, the equation of the line tangent to the curve at the given point is: