Question 4f(ii)

2 marks · average 0.5 72% scored 0

The question in our own words · page 25

From part i, the tangent at \(x=p\) is \(t(x)=\cos(p)(x-p)+\sin(p)+1\). As \(p\) moves between \(0\) and \(\tfrac{5\pi}{2}\), how low and how high can the tangent’s \(y\)-intercept go?

Open the VCAA paper ↗
1 / 1

More from Methods 2025 Exam 2

Every question in this paper with how students scored ›Next: Question 4f(iii) A sine curve and its tangents ›Tell me what’s confusing beta feedback ↗