The tangent to the curve y = x^2 - 5x + 5 whi | vedclass

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(B) Given curve is $y = x^2 - 5x + 5$. The slope of the tangent is given by $\frac{dy}{dx} = 2x - 5$. The given line is $2y = 4x + 1$,which can be wri

Vedclass · Accepted Answer

$\left( \frac{1}{8}, -7 \right)$

Vedclass · Answer

(B) Given curve is $y = x^2 - 5x + 5$. The slope of the tangent is given by $\frac{dy}{dx} = 2x - 5$. The given line is $2y = 4x + 1$,which can be written as $y = 2x + \frac{1}{2}$. The slope of this line is $2$. Since the tangent is parallel to the line,their slopes are equal: $2x - 5 = 2 \implies 2x = 7 \implies x = \frac{7}{2}$. Substituting $x = \frac{7}{2}$ into the curve equation: $y = \left( \frac{7}{2} \right)^2 - 5\left( \frac{7}{2} \right) + 5 = \frac{49}{4} - \frac{35}{2} + 5 = \frac{49 - 70 + 20}{4} = -\frac{1}{4}$. The point of tangency is $\left( \frac{7}{2}, -\frac{1}{4} \right)$. The equation of the tangent line is $y - y_1 = m(x - x_1)$,where $m = 2$: $y - (-\frac{1}{4}) = 2(x - \frac{7}{2}) \implies y + \frac{1}{4} = 2x - 7 \implies y = 2x - 7 - \frac{1}{4} \implies y = 2x - \frac{29}{4}$. Now,we check which point satisfies this equation. For option $B$: $x = \frac{1}{8}$,$y = 2(\frac{1}{8}) - \frac{29}{4} = \frac{1}{4} - \frac{29}{4} = -\frac{28}{4} = -7$. Thus,the tangent passes through the point $\left( \frac{1}{8}, -7 \right)$.

The tangent to the curve $y = x^2 - 5x + 5$ which is parallel to the line $2y = 4x + 1$ also passes through the point

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