Question

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1.
find an equation of the line that is tangent to the graph of $f$ and parallel to the given line.
function
$f(x) = x^3$
line
$12x - y + 1 = 0$
$y = 12x - 16$
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Explanation:

Step1: Find slope of given line

Rewrite line to slope-intercept form:
$12x - y + 1 = 0 \implies y = 12x + 1$
Slope $m = 12$

Step2: Find derivative of $f(x)$

Derivative gives tangent slope:
$f'(x) = \frac{d}{dx}(x^3) = 3x^2$

Step3: Solve for x-coordinate

Set derivative equal to slope:
$3x^2 = 12 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2$

Step4: Find y-coordinates on $f(x)$

For $x=2$: $f(2) = 2^3 = 8$
For $x=-2$: $f(-2) = (-2)^3 = -8$

Step5: Find tangent equations

Use point-slope form $y - y_1 = m(x - x_1)$:
For $(2, 8)$: $y - 8 = 12(x - 2) \implies y = 12x - 16$
For $(-2, -8)$: $y - (-8) = 12(x - (-2)) \implies y = 12x + 16$

Answer:

$y = 12x - 16$ and $y = 12x + 16$