Video: AQA GCSE Mathematics Higher Tier Pack 2 β€’ Paper 1 β€’ Question 23

A curve has the equation 𝑦 is equal to π‘₯ minus three all squared minus four. Find the coordinates of the turning point of the curve. Circle your answer. Is it three, negative four; three, four; negative three, four; negative three, negative four; or four, three?

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Video Transcript

A curve has the equation 𝑦 is equal to π‘₯ minus three all squared minus four. Find the coordinates of the turning point of the curve. Circle your answer. Is it three, negative four; three, four; negative three, four; negative three, negative four; or four, three?

This equation is actually in completed square form. And there is a formula we can use to help us find the coordinates of the turning point of the curve. But let’s first consider the transformation of a curve.

It’s a transformation of the graph 𝑓 of π‘₯ is equal to π‘₯ squared. Remember, this is a parabola with a turning point at the origin, at zero, zero. The graph of 𝑓 of π‘₯ minus three which in this case is π‘₯ minus three all squared is a translation by the vector three, zero.

The graph moves three unit in the positive π‘₯-direction. Finally, 𝑓 of π‘₯ minus three minus four is in this case π‘₯ minus three all squared minus four, which we’ll notice it’s the equation of the graph in our question. This time that’s a translation of the original by three, negative four. It still moves the original graph three units to the right, but it also moves it four units down and it takes our turning point to three, negative four.

And we’ve used transformations of graphs to find the turning point of the equation π‘₯ minus three all squared minus four. In fact, a graph with the equation π‘₯ plus π‘Ž all squared plus 𝑏 has a turning point at negative π‘Žπ‘. This comes from it being a translation of the graph 𝑦 is equal to π‘₯ squared by a vector of negative π‘Žπ‘.

In this case, once again, we can see that the turning point of the curve is three, negative four.

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