=4x3−6x2+2x+5 The slope of tangent line at (x1,y1) is m=4x13−6x12+2x1+5 The equation of tangent line is y−y1=(4x13−6x12+2x1+5)(x−x1) Since, the tangent line passes through (0,0) ⇒0−y1‌=(4x13−6x12+2x1+5)(0−x1) y1‌=(4x13−6x12+2x1+5)x1 Since, ( x1,y1 ) is on the curve y1=x14−2x13+x12+5x1 Substituting this into the previous equation, we get ‌x14−2x13+x12+5x1 ‌=(4x13−6x12+2x1+5)x1 ⇒x14−2x13+x12+5x1 ‌=4x14−6x13+2x12+5x1 ⇒3x14−4x13+x12=0 ⇒x12(3x12−4x1+1)=0 ⇒x12(3x1−1)(x1−1)=0 ⇒x1‌=0,x1=‌