The vertical displacement hhh, in millimetres, of a high-precision mechanical component is modeled by the function
h(t)=t2(t+a) h(t) = t^2(t + a) h(t)=t2(t+a)where ttt is the time in seconds and aaa is a positive constant.
Sketch the curve with equation y=t2(t+a)y = t^2(t + a)y=t2(t+a).
A second model for the displacement, H(t)H(t)H(t), includes a damping offset and is given by
H(t)=t2(t+a)+54 H(t) = t^2(t + a) + 54 H(t)=t2(t+a)+54(i) Given that t+6t + 6t+6 is a factor of the polynomial H(t)H(t)H(t), use the factor theorem to show that a=4.5a = 4.5a=4.5.
State the single transformation which maps the curve with equation y=t2(t+4.5)y = t^2(t + 4.5)y=t2(t+4.5) onto the curve with equation y=t2(t+4.5)+54y = t^2(t + 4.5) + 54y=t2(t+4.5)+54.
The expression t2(t+4.5)+54t^2(t + 4.5) + 54t2(t+4.5)+54 can be written in the form (t+6)(t2+bt+c)(t + 6)(t^2 + bt + c)(t+6)(t2+bt+c). Without finding the values of bbb and ccc, use your knowledge of the transformation in part (b)(ii) and the sketch in part (a) to explain why
b2<4c b^2 < 4c b2<4c216 exam-style questions on WJEC A Level Maths 3.2 Algebra and Functions (A-level only), covering 3.2.1 Algebra and Functions (A-level only), 3.2.2 Algebra and Functions (A-level only), 3.2.3 Algebra and Functions (A-level only), 3.2.4 Algebra and Functions (A-level only), 3.2.5 Algebra and Functions (A-level only), 3.2.6 Algebra and Functions (A-level only), and 3.2 Algebra and Functions (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.