Quick questions on Power, polynomial and circle graphs for HSC Maths Advanced: the shapes of $y = x^n$ for even and odd powers, sketching factored cubics and polynomials with a sign table, circles $x^2 + y^2 = r^2$ centred at the origin and shifted circles, and the rectangular hyperbola $y = \frac{1}{x}$ with its vertical and horizontal asymptotes, with worked examples and practice questions

4short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What is powers of x?

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The simplest non-linear graphs are the power functions

y = x^n

. Their shape depends almost entirely on whether the power

n

is even or odd, because that decides what happens to a negative input. A negative number raised to an even power is positive, while raised to an odd power it stays negative:

(-2)^2 = 4

but

(-2)^3 = -8

What is sketching a factored cubic with a sign table?

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A cubic is a polynomial of degree

3

, such as

y = x^3 - 2x^2 - 5x + 6

. You cannot sketch a general cubic by hand at this stage, but if it is factored into linear factors you can, because the factors hand you the zeros straight away. The graph can only change sign at a zero, so a sign table built by testing one value in each interval between the zeros tells you exactly where the curve is above the axis and where it is below.

What is circles centred at the origin?

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A circle is the set of points a fixed distance

r

(the radius) from a fixed centre. For a circle centred at the origin, a point

(x, y)

is on the circle exactly when its distance from

O

r

. By Pythagoras (the distance formula), that distance squared is

x^2 + y^2

, so the equation is simply

What is the rectangular hyperbola y = 1/x?

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The reciprocal function

y = \dfrac{1}{x}

(equivalently

xy = 1

) graphs as a rectangular hyperbola: a curve in two separate pieces, called branches. It is undefined at

x = 0

(you cannot divide by zero), so there is a gap there. A short table makes the shape clear: at

x = \tfrac12, 1, 2, 4

the values are

y = 2, 1, \tfrac12, \tfrac14

, a curve falling steeply then flattening; the negative inputs give the matching negative outputs

y = -2, -1, -\tfrac12, -\tfrac14

. So one branch sits in the first quadrant (both coordinates positive) and one in the third (both negative); the function is odd, with point symmetry about the origin.

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