Calculus, 2ADV C1 EO-Bank 3 MC

At which point on the curve \(y = x^{2}-6x + 8\) can a normal be drawn such that it is inclined at 45\(^{\circ}\) to the positive \(x\)-axis?

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\(C\)

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\(y = x^{2}-6x + 8\)

\(y^{′} = 2x-6\)

If the normal is inclined at \(45^{\circ}\) to the positive x-axis then:

\(m_{\text{normal}} = \tan 45^{\circ} = 1\)

\(\text{Since } m_{\text{tangent}} \times m_{\text{normal}} = -1,\)

\(\therefore m_{\text{tangent}} = -1.\)

Find \(x\) when \(y^{′} = -1:\)

Find \(y:\)

\(y\)	\(= \left(\dfrac{5}{2}\right)^{2}-6\left(\dfrac{5}{2}\right) + 8\)
	\(=\dfrac{25}{4}-15 + 8\)
	\(= -\dfrac{3}{4}\)

\(\Rightarrow C\)