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Blizzard [7]
3 years ago
10

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

Mathematics
1 answer:
Alexxandr [17]3 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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2x⁴+17x²+8

Let X = x²     →→(x = + or -√X )

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X' = [-b+√(b²-4ac)]/2a             and      X" = [-b-√(b²-4ac)]/2a   
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3rd ROOT: X" = + 2√2.i
4th ROOT: X" = - 2√2.i


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