How does the parabola \(y=x^2-2(k+2)x+(k+2)^2\) meet the \(x\)-axis?
Answer and explanation
Correct answer: It will only touch the axis
The equation can be written as \(y=(x-(k+2))^2\). On the \(x\)-axis, \(y=0\), so \((x-(k+2))^2=0\), which has the single real root \(x=k+2\). Therefore, for every real value of \(k\), the parabola only touches the \(x\)-axis; it is not restricted to \(k=-2\). Exam tip: when the discriminant of the corresponding quadratic is \(D=0\), the graph touches the axis at exactly one point.
Frequently asked questions
What is the correct answer to this question?
It will only touch the axis
Why is this the correct answer?
The equation can be written as \(y=(x-(k+2))^2\). On the \(x\)-axis, \(y=0\), so \((x-(k+2))^2=0\), which has the single real root \(x=k+2\). Therefore, for every real value of \(k\), the parabola only touches the \(x\)-axis; it is not restricted to \(k=-2\). Exam tip: when the discriminant of the corresponding quadratic is \(D=0\), the graph touches the axis at exactly one point.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
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