The curve with equation \(y = \mathrm{f}(x)\) for \(x \leq b\), where \(\mathrm{f}(x)\) is a quadratic function, intersects the \(x\)-axis at the points \((a,0)\) and \((b,0)\), and the \(y\)-axis at the point \((0,c)\). The graph of \(y = \mathrm{f}(x)\) is shown in Figure 1. The scales on the \(x\)- and \(y\)-axes are the same.
Figure 1
Sketch the graph of \(y = \mathrm{f}(\left| x \right|)\), labelling the points where the graph intersects or touches the axes.[2]
Sketch the graph of \(y = \frac{1}{\mathrm{f}(x)}\), labelling the points where the graph intersects or touches the axes, as well as the equations of any asymptotes.[2]
Describe fully a sequence of transformations which transforms the graph of \(y = \mathrm{f}(2x+1)\) onto the graph of \(y = \mathrm{f}(x)\).[2]
The function \(y = \mathrm{g}(x)\) is such that \(\mathrm{g}(x) = \mathrm{f}(x)\) for \(x \leq k\), and \(y = \mathrm{g}^{-1}(x)\) exists.
Find the largest possible value of \(k\) in terms of \(a\) and \(b\).[1]
Using the value of \(k\) found in (b)(i), sketch the graph of \(y = \mathrm{g}^{-1}(x)\) on Figure 1 in the Answer Booklet, labelling the intersections with the axes. Write down the range of \(\mathrm{g}^{-1}\) in terms of \(a\) and \(b\).[2]
Explain why the solution to \(\mathrm{g}^{-1}(x) = x\) satisfies the equation \(\mathrm{g}^{-1}(x) = \mathrm{g}(x)\).[1]
Video Solution:
Video solution locked
Solution:
Solution locked
Sign in to view the step-by-step solution
Similar questions are unavailable for this question.
Answer:(a)(i) Domain \([-b,b]\); intercepts \((-b,0)\), \((b,0)\) and \((0,c)\). (ii) Asymptotes \(x=a\), \(x=b\) and \(y=0\); intercept \(\left(0,\dfrac1c\right)\). (iii) Stretch parallel to the \(x\)-axis by factor \(2\), then translate \(1\) unit right. (b)(i) \(k=\dfrac{a+b}{2}\) (ii) Intercepts \((c,0)\) and \((0,a)\); range \(\left(-\infty,\dfrac{a+b}{2}\right]\).