Complete the table of values for \(y=100-\dfrac{120}{x^2}\).[1]
\(x\)
\(-2\)
\(-1\)
\(1\)
\(2\)
\(3\)
\(4\)
\(5\)
\(6\)
\(y\)
\(-20\)
\(-20\)
\(70\)
\(86.7\)
\(92.5\)
\(95.2\)
\(96.7\)
On the grid, draw the graph of \(y=100-\dfrac{120}{x^2}\) for \(-2\leq x\leq6\).[3]
By drawing a tangent, estimate the gradient of the curve at \(x=3.5\).[2]
Carl has two types of power banks, Power Bank A and Power Bank B, for charging his mobile phone.
Using Power Bank A, the battery percentage, \(y\%\), of Carl’s mobile phone after \(x\) hours of charging is given by the function \(y=100-\dfrac{120}{x^2},x\geq2\).
What does your answer to part (a)(iii) tell you about Power Bank A?[1]
The solutions to the equation \(x^3-5x^2+12=0\) represent the charging durations for which Power Bank A and Power Bank B charge Carl’s phone to the same battery percentage.
Carl has only 4 hours to charge his mobile phone.
By drawing a suitable straight line on the grid, determine whether Carl should use Power Bank A or Power Bank B.[4]
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Answer:(a)(i) 70 (ii) See the two-branch graph (iii) About 5.6. (b)(i) About 5.6 percentage points per hour at 3.5 hours. (ii) Power Bank A: 92.5% versus 90% after 4 hours, using the intended line \(y=10x+50\).