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2 years ago

8

Find an equation for the tangent to the curve at the given point y=x^3 , (2,8)

1 answer:

Yuliya22 [10]2 years ago

5 0

$\bf y=x^3\implies \left. \cfrac{dy}{dx}=3x^2 \right|_{x=2}\implies \stackrel{\stackrel{m}{\downarrow }}{12} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{8})~\hspace{10em} slope = m\implies 12 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-8=12(x-2) \\\\\\ y-8=12x-24\implies y=12x-16$

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Write an equation of the line perpendicular to the given line that contains P.

Yakvenalex [24]

Answer:

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Step-by-step explanation:

The slope of the given line is 3. The slope of the perpendicular line is the negative reciprocal of that, -1/3. Then the point-slope equation is ...

y -y1 = m(x -x1) . . . . line with slope m through point (x1, y1)

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Step-by-step explanation:

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A positive real number is 2 less than another. When 4 times the larger is added to the square of the smaller, the result is 49.

Kobotan [32]

Answer:

The numbers are

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Step-by-step explanation:

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x -----> the smaller positive real number

y -----> the larger positive real number

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A positive real number is 2 less than another

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When 4 times the larger is added to the square of the smaller, the result is 49

so

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substitute equation A in equation B

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solve for x

$x^2+4x-41=0$

The formula to solve a quadratic equation of the form

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$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^2+4x-41=0$

so

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substitute in the formula

$x=\frac{-4\pm\sqrt{4^{2}-4(1)(-41)}} {2(1)}$

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so

The positive real number is

$x=-2+3\sqrt{5}$

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$y=-2+3\sqrt{5}+2$

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