A designer fits a parabolic arch modelled by y = x^2 - 2mx + (m^2 + m - 6). The arch just touches the x-axis (one point only) for some m. Find the value of m for which the curve touches the x-axis exactly once.

Question

Accepted Answer

6. Touching once means equal roots, so D = 0: (-2m)^2 - 4(1)(m^2 + m - 6) = 0, i.e. 4m^2 - 4m^2 - 4m + 24 = 0, so -4m + 24 = 0, giving m = 6.

A designer fits a parabolic arch modelled by $y = x^{2} - 2 m x + (m^{2} + m - 6)$ . The arch just touches the $x$ -axis (one point only) for some $m$ .
Find the value of $m$ for which the curve touches the $x$ -axis exactly once.

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More Class 10 Discriminant and Nature of Roots practice

A designer fits a parabolic arch modelled by y=x2−2mx+(m2+m−6). The arch just touches the x-axis (one point only) for some m.Find the value of m for which the curve touches the x-axis exactly once.

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Discriminant and Nature of Roots practice

A designer fits a parabolic arch modelled by $y = x^{2} - 2 m x + (m^{2} + m - 6)$ . The arch just touches the $x$ -axis (one point only) for some $m$ .
Find the value of $m$ for which the curve touches the $x$ -axis exactly once.