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ACT Math : Algebra

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

Example Questions

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Example Question #8 : 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=2

y=0

y=x^2

y=2x

y=1

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 #9 : 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=3x^2

y=x^3+2

y=12x-14

y=12x

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 #10 : 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=x^2+3

y=2x

y=2

y=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=4

y=5x-4

y=5x

y=4x-5

y=2x+1

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

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Example Question #11 : How To Find The Equation Of A Tangent Line

What is the equation of a tangent line to

y=x^2+x^5

at the point

(1,2) ?

Possible Answers:

y=2x+5x^4

y=7x-5

y=2x

y=5x-7

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 #11 : Coordinate Plane

What is the equation of a tangent line to

y=x^2

at the point

(3,9) ?

Possible Answers:

y=6x-9

y=2x

y=9x-6

y=2

y=0

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

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Example Question #14 : Algebra

What is the equation of a tangent line to

y=x^4

at the point

(2,8) ?

Possible Answers:

y=x^4

y=4x^3

y=32

y=32x-56

y=56x-32

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 #15 : Algebra

What is the equation of a tagent line to

y=x^2

at the point

(-2,8) ?

Possible Answers:

y=2x

y=-4x

y=4x

y=-4

y=x^2

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

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Example Question #1 : Coordinate Geometry

There is a line defined by the equation below:

3x+4y=12

There is a second line that passes through the point (1,2) and is parallel to the line given above. What is the equation of this second line?

Possible Answers:

$y=-\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.625$

$y=\frac{3}{4}x+2.75$

$y=-\frac{3}{4}x+2.75$

$y=\frac{3}{4}x+1.25$

Correct answer:

$y=-\frac{3}{4}x+2.75$

Explanation:

Parallel lines have the same slope. Solve for the slope in the first line by converting the equation to slope-intercept form.

3x + 4y = 12

4y = –3x + 12

y = –(3/4)x + 3

slope = –3/4

We know that the second line will also have a slope of –3/4, and we are given the point (1,2). We can set up an equation in slope-intercept form and use these values to solve for the y-intercept.

y = mx + b

2 = –3/4(1) + b

2 = –3/4 + b

b = 2 + 3/4 = 2.75

Plug the y-intercept back into the equation to get our final answer.

y = –(3/4)x + 2.75

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Example Question #5 : Coordinate Geometry

What is the equation of a line that is parallel to 3x + 5y = 8 and passes through (10, 4) ?

Possible Answers:

$y=\frac{3}{4}x-2$

$y=\frac{5}{3}x+5$

$y=\frac{5}{6}x-10$

$y=\frac{3}{8}x-5$

$y=-\frac{3}{5}x+10$

Correct answer:

$y=-\frac{3}{5}x+10$

Explanation:

To solve, we will need to find the slope of the line. We know that it is parallel to the line given by the equation, meaning that the two lines will have equal slopes. Find the slope of the given line by converting the equation to slope-intercept form.

3x + 5y = 8

5y=-3x+8

$y=-\frac{3}{5}x+\frac{8}{5}$

The slope of the line will be $-\frac{3}{5}$ . In slope intercept-form, we know that the line will be $y=-\frac{3}{5}x+b$ . Now we can use the given point to find the y-intercept.

$y=-\frac{3}{5}x+b$

$4=-\frac{3}{5}(10)+b$

4=-6+b

10=b

The final equation for the line will be $y=-\frac{3}{5}x+10$ .

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ACT Math : Algebra

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #8 : How To Find The Equation Of A Tangent Line

Example Question #9 : How To Find The Equation Of A Tangent Line

Example Question #10 : How To Find The Equation Of A Tangent Line

Example Question #11 : Algebra

Example Question #11 : How To Find The Equation Of A Tangent Line

Example Question #11 : Coordinate Plane

Example Question #14 : Algebra

Example Question #15 : Algebra

Example Question #1 : Coordinate Geometry

Example Question #5 : Coordinate Geometry

All ACT Math Resources

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