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

y2(y2−4)=x2(x2−5)y2(y2−4)=x2(x2−5) Multiplying the polynomials gets us to y4−4y2=x4−5x2y4−4y2=x4−5x2. Taking the derivative with respect to xx gets us: 4y3y′−>!8yy′=4x3−10x4y3y′−>!8yy′=4x3−10x. Factoring to get y′y′ by itself: y′(4y3−8y)=4x3−10)y′(4y3−8y)=4x3−10). Divide through to get y′y′ by itself: y′=4x3−10x4y3−8yy′=4x3−10x4y3−8y. You could make your life a bit easier by factoring this into y′=2x(2x2−5)4y(y2−2)y′=2x(2x2−5)4y(y2−2). You could cancel out a factor of 22 to get y′=x(2x2−5)2y(y2−2)y′=x(2x2−5)2y(y2−2). To find the slope, plug in your points x=0,y=−2x=0,y=−2 into our equation for y′y′ to find the slope of the line. Note that the slope is 00. To find the equation of the tangent line, use that value for mm you just found (m=0m=0) and your given points into the point-slope formula and you find that the tangent line is y=−2y=−2.

Thats what my Aunt said... Idk