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Precalculus : Tangents To a Curve

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Example Questions

Example Question #2 : Find The Equation Of A Line Tangent To A Curve At A Given Point

Write the equation for the tangent line to $y=\frac{1}{4}x^3 +3x$ at x=2 .

Possible Answers:

y =9x-10

y=6x+4

$y=9x-11\frac{7}{8}$

$y=6x-5 \frac{7}{8}$

y=6x-4

Correct answer:

y=6x-4

Explanation:

First, find the slope of the tangent line by taking the first derivative:

$y'=\frac{3}{4}x ^2 + 3$

To finish determining the slope, plug in the x-value, 2:

$\frac{3}{4}(2)^2 +3 = \frac{3}{4} \cdot 4 + 3 = 3+3=6$ the slope is 6

Now find the y-coordinate where x is 2 by plugging in 2 to the original equation:

$y=\frac{1}{4} (2)^3 +3(2) = \frac{1}{4} \cdot 8 + 6 = 2+6 = 8$

To write the equation, start in point-slope form and then use algebra to get it into slope-intercept like the answer choices.

y-8=6(x-2) distribute the 6

y-8 = 6x-12 add 8 to both sides

y=6x-4

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Example Question #1 : Find The Equation Of A Line Tangent To A Curve At A Given Point

Write the equation for the tangent line for $y = \frac{1}{3} x^3 +2x^2 -x+7$ at x=-2 .

Possible Answers:

$y=-5x-4\frac{1}{3}$

$y = 5x + 4\frac{1}{3}$

$y=-3x+8 \frac{1}{3}$

$y = -5x+\frac{1}{3}$

$y=-5x+4 \frac{1}{3}$

Correct answer:

$y=-5x+4 \frac{1}{3}$

Explanation:

First, take the first derivative in order to find the slope:

y'=x^2 +4x-1

To continue finding the slope, plug in the x-value, -2:

(-2)^2 + 4(-2) - 1 = 4-8-1=-5

Then find the y-coordinate by plugging -2 into the original equation:

$\frac{1}{3}(-2)^2 +2(-2)^2 -(-2)+7 =\frac{1}{3}\cdot -8 + 8 + 2 + 7 = -2 \frac{2}{3} +17=14 \frac{1}{3}$

The y-coordinate is $14 \frac{1}{3}$

Now write the equation in point-slope form then algebraically manipulate it to match one of the slope-intercept forms of the answer choices.

$y-14 \frac{1}{3} = -5(x+2)$ distribute the -5

$y - 14\frac{1}{3} = -5 x -10$ add $14 \frac{1}{3}$ to both sides

$y = -5x +4 \frac{1}{3}$

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Example Question #1 : Find The Equation Of A Line Tangent To A Curve At A Given Point

Write the equation for the tangent line to $y = \frac{1}{4}(x^2 + 3x -10 )$ at x=-3 .

Possible Answers:

$y = \frac{ 3}{2}x + 2$

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

$y = x + \frac{1}{2}$

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

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

Correct answer:

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

Explanation:

First distribute the $\frac{1}{4}$ . That will make it easier to take the derivative:

$y = \frac{1}{4}x^2 + \frac{3}{4} x - \frac{10}{4}$

Now take the derivative of the equation:

$y' = \frac{2}{4} x + \frac{3}{4} = \frac{1}{2}x + \frac{3}{4}$

To find the slope, plug in the x-value -3:

$\frac{1}{2} (-3) + \frac{3}{4} = \frac{-3}{2}+\frac{3}{4}=-\frac{6}{4} + \frac{3}{4} =- \frac{3}{4}$

To find the y-coordinate of the point, plug in the x-value into the original equation:

$y=\frac{1}{4}( (-3)^2 +3(-3)-10) = \frac{1}{4} (9-9-10) = \frac{1}{4} (-10) =\frac{-10}{4}$

Now write the equation in point-slope, then use algebra to get it into slope-intercept like the answer choices:

$y +\frac{10}{4} = - \frac{3}{4} (x+3)$ distribute

$y + \frac{10}{4} = -\frac{3}{4} x - \frac{9}{4}$ subtract $\frac{10}{4}$ from both sides

$y =- \frac{3}{4}x - \frac{19}{4}$ write as a mixed number

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

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Precalculus : Tangents To a Curve

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #2 : Find The Equation Of A Line Tangent To A Curve At A Given Point

Example Question #1 : Find The Equation Of A Line Tangent To A Curve At A Given Point

Example Question #1 : Find The Equation Of A Line Tangent To A Curve At A Given Point

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