Assume that y=ax2+bx+cy=ax^2+bx+cy=ax2+bx+c
If x=0x=0x=0 then y=0y=0y=0 .We have 0=a⋅0+b⋅0+c→c=00=a\cdot0+b\cdot0+c\to c=00=a⋅0+b⋅0+c→c=0.
For point (6,4): 4=a⋅62+b⋅6→a=4−6b364=a\cdot6^2+b\cdot6\to a=\frac{4-6b}{36}4=a⋅62+b⋅6→a=364−6b
For point (10,4): 4=a⋅102+b⋅10→a=4−10b1004=a\cdot10^2+b\cdot10\to a=\frac{4-10b}{100}4=a⋅102+b⋅10→a=1004−10b
So,
4−6b36=4−10b100→b=1.067\frac{4-6b}{36}=\frac{4-10b}{100}\to b=1.067364−6b=1004−10b→b=1.067
At point (0,0)
dydx=2ax+1.067=2a⋅0+1.067=tan(α)\frac{dy}{dx}=2ax+1.067=2a\cdot0+1.067=\tan(\alpha)dxdy=2ax+1.067=2a⋅0+1.067=tan(α)
1.067=tan(α)→α=46,85°1.067=\tan(\alpha)\to \alpha=46,85°1.067=tan(α)→α=46,85°
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