ACT Math - Tangent Lines

What is the equation of a tangent line to

at point ?

Explanation

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find our slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}+4$

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug the point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation becomes,

Once we rearrange, the equation is

Find the slope of the tangent line to where .

Explanation

To find the slope of the tangent line, we must take the derivative.

By using the Power Rule we will be able to find the derivative:

$f(x)=a^n \rightarrow f'(x)=na^{n-1}$

Therefore derivative of is .

Now we plug in , giving us .

What is the equation of a tangent line to

at point ?

Explanation

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find our slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}+4$

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug the point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation becomes,

Once we rearrange, the equation is

Find the slope of the tangent line to where .

Explanation

To find the slope of the tangent line, we must take the derivative.

By using the Power Rule we will be able to find the derivative:

$f(x)=a^n \rightarrow f'(x)=na^{n-1}$

Therefore derivative of is .

Now we plug in , giving us .

Find the tangent line equation to

at point

Explanation

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find our slope at our $x_{1}$ by plugging in the value into our derivative equation and solving.

Thus, the slope is

$m=2x_{1}$

To find the equation of a tangent line of a given point $(x_{1},y_{1})$

We plug into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Find the tangent line equation to

at point

Explanation

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find our slope at our $x_{1}$ by plugging in the value into our derivative equation and solving.

Thus, the slope is

$m=2x_{1}$

To find the equation of a tangent line of a given point $(x_{1},y_{1})$

We plug into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

What is the tangent line equation of

at point

Explanation

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find the slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=4x_{1}^3+4$

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug it into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

What is the tangent line equation of

at point

Explanation

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find the slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=4x_{1}^3+4$

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug it into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Find the equation of a tangent line to

for the point

Explanation

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find the slope by plugging in our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}$

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug our point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Find the equation of a tangent line to

for the point

Explanation

To find an equation tangent to

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

due to power rule $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find the slope by plugging in our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}$

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug our point into

$y-y_{1}=m(x-x_{1})$ .

Therefore our equation is

Once we rearrange, the equation is

Tangent Lines

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation