Question 1075331
y"=12x^2-8
y'=4x^3-8x+C1
y=x^4-4x^2+C1x+C2
When x=4 y=256-64+4C1+C2
4C1+C2=0
The slope of the line tangent to this is perpendicular to 224y=-x+448; y=-x/224+2, and it is 224, the negative reciprocal
point-slope y-y1=m(x-x1) m slope (x1,y1) point
y-192=224(x-4)
y=224x-704
The slope of 224=4x^3-8x+C1, where x=4
224=256-32+C1, so C1=0, C2=0
The curve is x^4-4x^2, or x^2(x+2)(x-2)
{{{graph(300,300,-10,10,-10,250,x^4-4x^2,224x-704,-(x/224)+2)}}}