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ACT Math : How to find the equation of a tangent line

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Example Questions

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

Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?

Possible Answers:

–3x + 4y = 1

3x – 4y = –1

3x – 4y = –25

3x + 4y = 7

Correct answer:

3x – 4y = –25

Explanation:

The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by Actmath_7_113_q7

The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.

The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.

The equation of the line is y – 4 = (3/4)(x – (–3))

Rearranging gives us: 3x – 4y = -25

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Example Question #1 : Algebra

Give the equation, in slope-intercept form, of the line tangent to the circle of the equation

$x^{2}+ y^{2} = 225$

at the point (12, -9) .

Possible Answers:

$y = \frac{4}{3} x- 25$

$y =-\frac{4}{3} x+7$

$y = \frac{3} {4}x-18$

None of the other responses gives the correct answer.

$y =- \frac{3} {4}x$

Correct answer:

$y = \frac{4}{3} x- 25$

Explanation:

The graph of the equation $x^{2}+ y^{2} = 225$ is a circle with center (0,0) .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints (0,0) and (12, -9) will have slope

$\frac{ y_{2} - y_{1} }{x_{2} - x_{1} } =\frac{-9 -0}{12-0} =\frac{-9 }{12 }=- \frac{3}{4}$ ,

so the tangent line has the opposite of the reciprocal of this, or $\frac{4}{3}$ , as its slope.

The tangent line therefore has equation

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

$y - (-9)= \frac{4}{3} (x-12 )$

$y +9= \frac{4}{3} x- 16$

$y = \frac{4}{3} x- 25$

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Example Question #2 : Algebra

Give the equation, in slope-intercept form, of the line tangent to the circle of the equation

$x^{2}-6x + y^{2} - 91 = 0$

at the point (-3, 8) .

Possible Answers:

$y = \frac{4} {3}x+12$

$y =- \frac{3}{4} x+\frac{23}{4}$

None of the other responses gives the correct answer.

$y = \frac{3}{4} x+\frac{41}{4}$

$y =- \frac{4} {3}x+4$

Correct answer:

$y = \frac{3}{4} x+\frac{41}{4}$

Explanation:

Rewrite the equation of the circle in standard form to find its center:

$x^{2}-6x + y^{2} - 91 = 0$

$x^{2}-6x + y^{2} = 91$

Complete the square:

$x^{2}-6x+9 + y^{2} = 91+9$

$\left (x-3 \right )^{2} + y^{2} = 100$

The center is (3,0) .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints (3,0) and (-3, 8) will have slope

$\frac{ y_{2} - y_{1} }{x_{2} - x_{1} } = \frac{8 -0}{-3-3} = \frac{8}{-6} = - \frac{4}{3}$ ,

so the tangent line has the opposite of the reciprocal of this, or $\frac{3}{4}$ , as its slope.

The tangent line therefore has equation

$y - 8= \frac{3}{4} (x- (-3))$

$y - 8= \frac{3}{4} (x+3)$

$y - 8= \frac{3}{4} x+\frac{9}{4}$

$y = \frac{3}{4} x+\frac{41}{4}$

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Example Question #3 : How To Find The Equation Of A Tangent Line

What is the equation of a tangent line to

y=x^2+4x+2

at point (1,7) ?

Possible Answers:

y=3x+1

y=x+6

y=6x+1

y=4x+1

y=x^2+4x+2

Correct answer:

y=6x+1

Explanation:

To find an equation tangent to

y=x^2+4x+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+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

$m=2x_{1}+4$

m=6 .

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,

y-7=6(x-1)

Once we rearrange, the equation is

y=6x+1

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Example Question #2 : Algebra

What is the tangent line equation of

y=x^4+4x

at point

(1,5) ?

Possible Answers:

y=4x^3+4

y=8x-3

y=8x+5

y=4

y=8x+8

Correct answer:

y=8x-3

Explanation:

To find an equation tangent to

y=x^4+4x

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+4

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

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

m=8 .

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

y-5=8(x-1)

Once we rearrange, the equation is

y=8x-3

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Example Question #4 : Lines

Find the equation of a tangent line to

y=x^2

for the point

(0,0) ?

Possible Answers:

y=2

y=2x

y=0

y=1

y=x^2

Correct answer:

y=0

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 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-0=0(x-0)

Once we rearrange, the equation is

y=0

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Example Question #3 : How To Find The Equation Of A Tangent Line

What is the equation of a tangent line to

y=5x^5+2

at the point

(1,7) ?

Possible Answers:

y=27x

y=25x-18

y=18x-25

y=25x^4

y=25x^4+2

Correct answer:

y=25x-18

Explanation:

To find an equation tangent to

y=5x^5+2

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

So,

y'=25x^4

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

$m=25x_{1}^4$

m=25 .

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-7=25(x-1)

Once we rearrange, the equation is

y=25x-18

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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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ACT Math : How to find the equation of a tangent line

Study concepts, example questions & explanations for ACT Math

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Example Questions

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

Example Question #1 : Algebra

Example Question #2 : Algebra

Example Question #3 : How To Find The Equation Of A Tangent Line

Example Question #2 : Algebra

Example Question #4 : Lines

Example Question #3 : How To Find The Equation Of A Tangent Line

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

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