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 : Coordinate Plane

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 = 7

–3x + 4y = 1

3x – 4y = –25

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

 

 

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

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

\(\displaystyle x^{2}+ y^{2} = 225\)

at the point \(\displaystyle (12, -9)\).

Possible Answers:

\(\displaystyle y = \frac{3} {4}x-18\)

\(\displaystyle y = \frac{4}{3} x- 25\)

None of the other responses gives the correct answer.

\(\displaystyle y =-\frac{4}{3} x+7\)

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

Correct answer:

\(\displaystyle y = \frac{4}{3} x- 25\)

Explanation:

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

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

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

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

The tangent line therefore has equation

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

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

\(\displaystyle y +9= \frac{4}{3} x- 16\)

\(\displaystyle y = \frac{4}{3} x- 25\)

Example Question #2 : Algebra

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

\(\displaystyle x^{2}-6x + y^{2} - 91 = 0\)

at the point \(\displaystyle (-3, 8)\).

Possible Answers:

\(\displaystyle y = \frac{4} {3}x+12\)

\(\displaystyle y =- \frac{3}{4} x+\frac{23}{4}\)

None of the other responses gives the correct answer.

\(\displaystyle y = \frac{3}{4} x+\frac{41}{4}\)

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

Correct answer:

\(\displaystyle y = \frac{3}{4} x+\frac{41}{4}\)

Explanation:

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

\(\displaystyle x^{2}-6x + y^{2} - 91 = 0\)

\(\displaystyle x^{2}-6x + y^{2} = 91\)

Complete the square:

\(\displaystyle x^{2}-6x+9 + y^{2} = 91+9\)

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

The center is \(\displaystyle (3,0)\)

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

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

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

The tangent line therefore has equation

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

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

\(\displaystyle y - 8= \frac{3}{4} x+\frac{9}{4}\)

\(\displaystyle y = \frac{3}{4} x+\frac{41}{4}\)

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

What is the equation of a tangent line to

\(\displaystyle y=x^2+4x+2\)

at point \(\displaystyle (1,7)\) ?

Possible Answers:

\(\displaystyle y=3x+1\)

\(\displaystyle y=x+6\)

\(\displaystyle y=6x+1\)

\(\displaystyle y=4x+1\)

\(\displaystyle y=x^2+4x+2\)

Correct answer:

\(\displaystyle y=6x+1\)

Explanation:

To find an equation tangent to

\(\displaystyle y=x^2+4x+2\)

we need to find the first derviative of this equation with respect to \(\displaystyle x\) to get the slope \(\displaystyle m\) of the tangent line.

So,

\(\displaystyle y'=2x+4\)

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

 

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

Thus, the slope is

\(\displaystyle m=2x_{1}+4\)

\(\displaystyle m=6\).

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

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

 

Therefore our equation becomes,

\(\displaystyle y-7=6(x-1)\)

Once we rearrange, the equation is

\(\displaystyle y=6x+1\)

 

Example Question #2 : Algebra

What is the tangent line equation of

\(\displaystyle y=x^4+4x\)

at point

\(\displaystyle (1,5)\) ?

Possible Answers:

\(\displaystyle y=4x^3+4\)

\(\displaystyle y=8x-3\)

\(\displaystyle y=8x+5\)

\(\displaystyle y=4\)

\(\displaystyle y=8x+8\)

Correct answer:

\(\displaystyle y=8x-3\)

Explanation:

To find an equation tangent to

\(\displaystyle y=x^4+4x\)

we need to find the first derviative of this equation with respect to \(\displaystyle x\) to get the slope \(\displaystyle m\) of the tangent line.

So,

\(\displaystyle y'=4x^3+4\)

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

 

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

Thus, the slope is

\(\displaystyle m=4x_{1}^3+4\)

\(\displaystyle m=8\).

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

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

 

Therefore our equation is

\(\displaystyle y-5=8(x-1)\)

Once we rearrange, the equation is

\(\displaystyle y=8x-3\)

Example Question #4 : Lines

Find the equation of a tangent line to

\(\displaystyle y=x^2\)

for the point

\(\displaystyle (0,0)\) ?

Possible Answers:

\(\displaystyle y=2\)

\(\displaystyle y=2x\)

\(\displaystyle y=0\)

\(\displaystyle y=1\)

\(\displaystyle y=x^2\)

Correct answer:

\(\displaystyle y=0\)

Explanation:

To find an equation tangent to

\(\displaystyle y=x^2\)

we need to find the first derviative of this equation with respect to \(\displaystyle x\) to get the slope \(\displaystyle m\) of the tangent line.

So,

\(\displaystyle y'=2x\)

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

 

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

Thus, the slope is

\(\displaystyle m=2x_{1}\)

\(\displaystyle m=0\).

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

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

 

Therefore our equation is

\(\displaystyle y-0=0(x-0)\)

Once we rearrange, the equation is

\(\displaystyle y=0\)

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

What is the equation of a tangent line to

\(\displaystyle y=5x^5+2\)

at the point

\(\displaystyle (1,7)\) ?

Possible Answers:

\(\displaystyle y=27x\)

\(\displaystyle y=25x-18\)

\(\displaystyle y=18x-25\)

\(\displaystyle y=25x^4\)

\(\displaystyle y=25x^4+2\)

Correct answer:

\(\displaystyle y=25x-18\)

Explanation:

To find an equation tangent to

\(\displaystyle y=5x^5+2\)

we need to find the first derviative of this equation with respect to \(\displaystyle x\) to get the slope \(\displaystyle m\) of the tangent line.

So,

\(\displaystyle y'=25x^4\)

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

 

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

Thus, the slope is

\(\displaystyle m=25x_{1}^4\)

\(\displaystyle m=25\).

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

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

 

Therefore our equation is

\(\displaystyle y-7=25(x-1)\)

Once we rearrange, the equation is

\(\displaystyle y=25x-18\)

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

Find the equation of a tangent line to

\(\displaystyle y=x^2+1\)

at the point

\(\displaystyle (0,1)\) ?

Possible Answers:

\(\displaystyle y=0\)

\(\displaystyle y=1\)

\(\displaystyle y=2\)

\(\displaystyle y=2x\)

\(\displaystyle y=x^2\)

Correct answer:

\(\displaystyle y=1\)

Explanation:

To find an equation tangent to

\(\displaystyle y=x^2+1\)

we need to find the first derviative of this equation with respect to \(\displaystyle x\) to get the slope \(\displaystyle m\) of the tangent line.

So,

\(\displaystyle y'=2x\)

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

 

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

Thus, the slope is

\(\displaystyle m=2x_{1}\)

\(\displaystyle m=0\).

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

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

 

Therefore our equation is

\(\displaystyle y-1=0(x-0)\)

Once we rearrange, the equation is

\(\displaystyle y=1\)

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

What is the equation of a tangent line to

\(\displaystyle y=x^3+2\)

at point

\(\displaystyle (2,10)\) ?

Possible Answers:

\(\displaystyle y=14x-12\)

\(\displaystyle y=x^3+2\)

\(\displaystyle y=3x^2\)

\(\displaystyle y=12x\)

\(\displaystyle y=12x-14\)

Correct answer:

\(\displaystyle y=12x-14\)

Explanation:

To find an equation tangent to

\(\displaystyle y=x^3+2\)

we need to find the first derviative of this equation with respect to \(\displaystyle x\) to get the slope \(\displaystyle m\) of the tangent line.

So,

\(\displaystyle y'=3x^2\)

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

 

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

Thus, the slope is

\(\displaystyle m=3x_{1}^2\)

\(\displaystyle m=12\).

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

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

 

Therefore our equation is

\(\displaystyle y-10=12(x-2)\)

Once we rearrange, the equation is

\(\displaystyle y=12x-14\)

 

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

Find the tangent line equation to

\(\displaystyle y=x^2+3\)

at point

\(\displaystyle (0,3)\) ?

Possible Answers:

\(\displaystyle y=2\)

\(\displaystyle y=2x\)

\(\displaystyle y=3\)

\(\displaystyle y=x^2+3\)

\(\displaystyle y=3x\)

Correct answer:

\(\displaystyle y=3\)

Explanation:

To find an equation tangent to

\(\displaystyle y=x^2+3\)

we need to find the first derviative of this equation with respect to \(\displaystyle x\) to get the slope \(\displaystyle m\) of the tangent line.

So,

\(\displaystyle y'=2x\)

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

 

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

Thus, the slope is

\(\displaystyle m=2x_{1}\)

\(\displaystyle m=0\).

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

We plug into

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

 

Therefore our equation is

\(\displaystyle y-3=0(x-0)\)

Once we rearrange, the equation is

\(\displaystyle y=3\)

 

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