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ACT Math : Tangent Lines

Study concepts, example questions & explanations for ACT Math

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14 Diagnostic Tests 767 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : How To Find The Equation Of A Tangent Line

Find the equation of a tangent line to

y=x^2+1

at the point

(0,1) ?

Possible Answers:

y=0

y=1

y=2

y=2x

y=x^2

Correct answer:

y=1

Explanation:

To find an equation tangent to

y=x^2+1

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

So,

y'=2x

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

$m=2x_{1}$

m=0 .

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

y-1=0(x-0)

Once we rearrange, the equation is

y=1

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Example Question #2 : How To Find The Equation Of A Tangent Line

What is the equation of a tangent line to

y=x^3+2

at point

(2,10) ?

Possible Answers:

y=14x-12

y=x^3+2

y=3x^2

y=12x

y=12x-14

Correct answer:

y=12x-14

Explanation:

To find an equation tangent to

y=x^3+2

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

So,

y'=3x^2

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

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

Thus, the slope is

$m=3x_{1}^2$

m=12 .

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 is

y-10=12(x-2)

Once we rearrange, the equation is

y=12x-14

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Example Question #1 : How To Find The Equation Of A Tangent Line

Find the tangent line equation to

y=x^2+3

at point

(0,3) ?

Possible Answers:

y=2

y=2x

y=3

y=x^2+3

y=3x

Correct answer:

y=3

Explanation:

To find an equation tangent to

y=x^2+3

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

So,

y'=2x

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

$m=2x_{1}$

m=0 .

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

y-3=0(x-0)

Once we rearrange, the equation is

y=3

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Example Question #11 : Algebra

What is the equation of a tangent line to

y=x^2+x

at the point

(2,6) ?

Possible Answers:

y=4x-5

y=2x+1

y=5x-4

y=5x

y=4

Correct answer:

y=5x-4

Explanation:

To find an equation tangent to

y=x^2+x

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

So,

y'=2x+1

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

$m=2x_{1}+1$

m=5 .

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

y-6=5(x-2)

Once we rearrange, the equation is

y=5x-4

Report an Error

Example Question #12 : Algebra

What is the equation of a tangent line to

y=x^2+x^5

at the point

(1,2) ?

Possible Answers:

y=5x-7

y=7x-5

y=2x

y=2x+5x^4

y=7x

Correct answer:

y=7x-5

Explanation:

To find an equation tangent to

y=x^2+x^5

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

So,

y'=2x+5x^4

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

$m=2x_{1}+5x_{1}^4$

m=7 .

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

y-2=7(x-1)

Once we rearrange, the equation is

y=7x-5

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Example Question #13 : Algebra

What is the equation of a tangent line to

y=x^2

at the point

(3,9) ?

Possible Answers:

y=6x-9

y=9x-6

y=0

y=2

y=2x

Correct answer:

y=6x-9

Explanation:

To find an equation tangent to

y=x^2

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

So,

y'=2x

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

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

Thus, the slope is

$m=2x_{1}$

m=6 .

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

y-9=6(x-3)

Once we rearrange, the equation is

y=6x-9

Report an Error

Example Question #642 : Act Math

What is the equation of a tangent line to

y=x^4

at the point

(2,8) ?

Possible Answers:

y=32

y=x^4

y=4x^3

y=56x-32

y=32x-56

Correct answer:

y=32x-56

Explanation:

To find an equation tangent to

y=x^4

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

So,

y'=4x^3

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

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

Thus, the slope is

$m=4x_{1}^3$

m=32 .

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

y-8=32(x-2)

Once we rearrange, the equation is

y=32x-56

Report an Error

Example Question #11 : Algebra

What is the equation of a tagent line to

y=x^2

at the point

(-2,8) ?

Possible Answers:

y=4x

y=-4x

y=-4

y=x^2

y=2x

Correct answer:

y=-4x

Explanation:

To find an equation tangent to

y=x^2

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

So,

y'=2x

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

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

Thus, the slope is

$m=2x_{1}$

m=-4 .

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

y-8=-4(x+2)

Once we rearrange, the equation is

y=-4x

Report an Error

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ACT Math : Tangent Lines

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Find The Equation Of A Tangent Line

Example Question #2 : How To Find The Equation Of A Tangent Line

Example Question #1 : How To Find The Equation Of A Tangent Line

Example Question #11 : Algebra

Example Question #12 : Algebra

Example Question #13 : Algebra

Example Question #642 : Act Math

Example Question #11 : Algebra

All ACT Math Resources

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