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2 years ago

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Find an equation for the line tangent to the curve y=x^3-4x+8 at the point (4,56)

1 answer:

Alexxandr [17]2 years ago

4 0

<span><span><span>y2</span>(<span>y2</span>−4)=<span>x2</span>(<span>x2</span>−5)</span><span><span>y2</span>(<span>y2</span>−4)=<span>x2</span>(<span>x2</span>−5)</span></span><span> Multiplying the polynomials gets us to </span><span><span><span>y4</span>−4<span>y2</span>=<span>x4</span>−5<span>x2</span></span><span><span>y4</span>−4<span>y2</span>=<span>x4</span>−5<span>x2</span></span></span><span>. Taking the derivative with respect to </span><span>xx</span><span> gets us: </span><span><span>4<span>y3</span><span>y′</span>−>!8y<span>y′</span>=4<span>x3</span>−10x</span><span>4<span>y3</span><span>y′</span>−>!8y<span>y′</span>=4<span>x3</span>−10x</span></span><span>. Factoring to get </span><span><span>y′</span><span>y′</span></span><span> by itself: </span><span><span><span>y′</span>(4<span>y3</span>−8y)=4<span>x3</span>−10)</span><span><span>y′</span>(4<span>y3</span>−8y)=4<span>x3</span>−10)</span></span><span>. Divide through to get </span><span><span>y′</span><span>y′</span></span><span> by itself: </span><span><span><span>y′</span>=<span><span>4<span>x3</span>−10x</span><span>4<span>y3</span>−8y</span></span></span><span><span>y′</span>=<span><span>4<span>x3</span>−10x</span><span>4<span>y3</span>−8y</span></span></span></span><span>. You could make your life a bit easier by factoring this into </span><span><span><span>y′</span>=<span><span>2x(2<span>x2</span>−5)</span><span>4y(<span>y2</span>−2)</span></span></span><span><span>y′</span>=<span><span>2x(2<span>x2</span>−5)</span><span>4y(<span>y2</span>−2)</span></span></span></span><span>. You could cancel out a factor of </span><span>22</span><span> to get </span><span><span><span>y′</span>=<span><span>x(2<span>x2</span>−5)</span><span>2y(<span>y2</span>−2)</span></span></span><span><span>y′</span>=<span><span>x(2<span>x2</span>−5)</span><span>2y(<span>y2</span>−2)</span></span></span></span><span>. To find the slope, plug in your points </span><span><span>x=0,y=−2</span><span>x=0,y=−2</span></span><span> into our equation for </span><span><span>y′</span><span>y′</span></span><span> to find the slope of the line. Note that the slope is </span><span>00</span><span>. To find the </span>equation<span> of the tangent line, use that value for </span><span>mm</span><span> you just found (</span><span><span>m=0</span><span>m=0</span></span><span>) and your given points into the point-slope formula and you find that the tangent line is </span><span><span>y=−2</span><span>y=−2</span></span><span>.

Thats what my Aunt said... Idk</span>

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What is -3(-6+8)^3-2(1-3)^3

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Assignment: $\bold{Solve \ Equation: \ -3\left(-6+8\right)^3-2\left(1-3\right)^3}$

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Answer: $\boxed{\bold{-8}}$

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Explanation: $\downarrow\downarrow\downarrow$

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[ Step One ] Follow PEMDAS Order Of Operations; Calculate Within Parenthesis

Note: $\bold{PEMDAS: \ Parenthesis, \ Exponents, \ Multiply, \ Divide, \ Add, \ Subtract}$

$\bold{-6+8: \ 2}$

[ Step Two ] Rewrite Equation

$\bold{-3\cdot \:2^3-2\left(1-3\right)^3}$

[ Step Three ] Calculate Within Parenthesis

$\bold{1-3: \ -2}$

[ Step Four ] Rewrite Equation

$\bold{-3\cdot \:2^3-2\left(-2\right)^3}$

[ Step Five ] Calculate Exponents

$\bold{2^3: \ 8}$

[ Step Six ] Rewrite Equation

$\bold{-3\cdot \:8-2\left(-8\right)}$

[ Step Seven ] Multiply

$\bold{2\left(-8\right): \ -16}$

[ Step Eight ] Rewrite Equation

$\bold{-24-\left(-16\right)}$

[ Step Nine ] Subtract

$\bold{-24-\left(-16\right): \ -8}$

[ Step Ten ] Rewrite Equation

$\bold{-8}$

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$\bold{\rightarrow Mordancy \leftarrow}$

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RideAnS [48]

Answer:

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Step-by-step explanation:

Kinetic energy, k = 1/2 mv^2

Where,

m = mass of the object in kilogram

v = velocity in meters per seconds

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48,000 = 1/2 × m × 20^2

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Using m = 240 kg

k = 1/2 mv^2

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