The normals to the curve y=x^{2}-x+1, drawn a | vedclass

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(C) Given the curve $y = x^2 - x + 1$.The slope of the tangent is $\frac{dy}{dx} = 2x - 1$.The slope of the normal is $m = -\frac{1}{dy/dx} = -\frac{1

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concurrent

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(C) Given the curve $y = x^2 - x + 1$.The slope of the tangent is $\frac{dy}{dx} = 2x - 1$.The slope of the normal is $m = -\frac{1}{dy/dx} = -\frac{1}{2x-1} = \frac{1}{1-2x}$.$1$. At $x_1 = 0$, $y_1 = 1$. Slope $m_1 = \frac{1}{1-0} = 1$.Equation: $y - 1 = 1(x - 0) \Rightarrow x - y + 1 = 0$ $(i)$.$2$. At $x_2 = -1$, $y_2 = (-1)^2 - (-1) + 1 = 3$. Slope $m_2 = \frac{1}{1-2(-1)} = \frac{1}{3}$.Equation: $y - 3 = \frac{1}{3}(x + 1) \Rightarrow 3y - 9 = x + 1 \Rightarrow x - 3y + 10 = 0$ (ii).$3$. At $x_3 = 5/2$, $y_3 = (5/2)^2 - 5/2 + 1 = 25/4 - 10/4 + 4/4 = 19/4$. Slope $m_3 = \frac{1}{1-2(5/2)} = \frac{1}{1-5} = -\frac{1}{4}$.Equation: $y - 19/4 = -\frac{1}{4}(x - 5/2) \Rightarrow 4y - 19 = -x + 5/2 \Rightarrow 2x + 8y - 43 = 0$ (iii).Solving $(i)$ and (ii): $x - y = -1$ and $x - 3y = -10$. Subtracting gives $2y = 9 \Rightarrow y = 9/2$. Then $x = y - 1 = 7/2$.Intersection point is $(7/2, 9/2)$.Check if $(7/2, 9/2)$ satisfies (iii): $2(7/2) + 8(9/2) - 43 = 7 + 36 - 43 = 43 - 43 = 0$.Since the point satisfies all three equations, the normals are concurrent.

The normals to the curve $y=x^{2}-x+1$, drawn at the points with the abscissae $x_{1}=0, x_{2}=-1$ and $x_{3}=5/2$, are:

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