The equation of a curve is \( y=kx^2-5x-6 \), and the equation of a line is \( y=3x-7k \).
Find the set of values of the constant \(k\) for which the line intersects the curve.
Solution
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At points of intersection,
\( kx^2-5x-6=3x-7k \).
So
\( kx^2-8x+7k-6=0 \).
For the line to intersect the curve, this equation must have real roots, so the discriminant is non-negative:
\( (-8)^2-4(k)(7k-6)\geq 0 \).
\( 64-28k^2+24k\geq 0 \).
\( 7k^2-6k-16\leq 0 \).
Factorise:
\( (7k+8)(k-2)\leq 0 \).
Answer: \( -\frac87\leq k\leq 2 \).
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