If the tangent drawn at the point (x_1, y_1), | vedclass

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(D) Given the curve $y = f(x) = x^4 - 2x^3 + x^2 + 5x$. The derivative is $f'(x) = 4x^3 - 6x^2 + 2x + 5$. The slope of the tangent at $(x_1, y_1)$ is

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$6$

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(D) Given the curve $y = f(x) = x^4 - 2x^3 + x^2 + 5x$. The derivative is $f'(x) = 4x^3 - 6x^2 + 2x + 5$. The slope of the tangent at $(x_1, y_1)$ is $m = 4x_1^3 - 6x_1^2 + 2x_1 + 5$. The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$. Since the tangent passes through the origin $(0, 0)$,we have $-y_1 = m(-x_1)$,which implies $y_1 = m x_1$. Substituting $y_1 = x_1^4 - 2x_1^3 + x_1^2 + 5x_1$ and $m = 4x_1^3 - 6x_1^2 + 2x_1 + 5$: $x_1^4 - 2x_1^3 + x_1^2 + 5x_1 = x_1(4x_1^3 - 6x_1^2 + 2x_1 + 5)$. $x_1^4 - 2x_1^3 + x_1^2 + 5x_1 = 4x_1^4 - 6x_1^3 + 2x_1^2 + 5x_1$. Rearranging terms: $3x_1^4 - 4x_1^3 + x_1^2 = 0$. Since $x_1 \in \mathbb{N}$,$x_1 
eq 0$,so we can divide by $x_1^2$: $3x_1^2 - 4x_1 + 1 = 0$. Solving the quadratic equation: $(3x_1 - 1)(x_1 - 1) = 0$. This gives $x_1 = 1$ or $x_1 = 1/3$. Since $x_1 \in \mathbb{N}$,we must have $x_1 = 1$. Then $y_1 = 1^4 - 2(1)^3 + 1^2 + 5(1) = 1 - 2 + 1 + 5 = 5$. Thus,$x_1 + y_1 = 1 + 5 = 6$.

If the tangent drawn at the point $(x_1, y_1)$,where $x_1, y_1 \in \mathbb{N}$,on the curve $y = x^4 - 2x^3 + x^2 + 5x$ passes through the origin,then $x_1 + y_1 =$

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