Powers, square roots, cubics and circles

Graphs of $y = x^n$

The shape of $y = x^n$ depends on whether $n$ is even or odd. Every one of them passes through $(0, 0)$ and $(1, 1)$.

Orange: y = xⁿ. Faint blue: y = x² for comparison.

Try this

• Between $x = -1$ and $x = 1$, which powers are flatter: higher or lower?
• Outside that interval, which grow faster?

The square root function

$y = \sqrt{x}$ is the top half of the sideways parabola $x = y^2$. Since $\sqrt{x}$ means the non-negative root, the domain is $x \ge 0$ and the range is $y \ge 0$.

Cubics from factored form

A cubic $y = a(x - \alpha)(x - \beta)(x - \gamma)$ crosses the $x$-axis at $\alpha$, $\beta$ and $\gamma$. To sketch it:

1. Mark the zeroes and the $y$-intercept.
2. Use the leading coefficient for the ends: if $a > 0$, the graph starts bottom-left and ends top-right. If $a < 0$, it is the other way round.
3. At a single zero the curve crosses the axis. At a double zero it touches and turns. At a triple zero it flattens out as it crosses (a horizontal point of inflection).

Try this

• Make two of the zeroes equal. What does the graph do there?
• Make all three equal. Describe the shape.
• Change the sign of $a$. What happens to the ends?

Example 1●●

Sketch $y = -x(x - 3)^2$.

Zeroes: $x = 0$ (crosses) and $x = 3$ (touches).

x = 0 is a single zero and x = 3 is a double zero.

Leading term $-x^3$: starts top-left, ends bottom-right.

Expanding the highest powers gives −x³, so a is negative.

Test a point between the zeroes: $x = 1$ gives $y = -1(4) = -4$, so the curve is below the axis there.

Sketch: come down from the top-left, cross at $0$, dip below, rise back to touch the axis at $3$, then head down.

Circles and semicircles

A circle fails the vertical line test, so it is not a function. Solving for $y$ gives two semicircles, and each of those is a function:

$$y = k + \sqrt{r^2 - (x-h)^2} \;\;\text{(upper)} \qquad y = k - \sqrt{r^2 - (x-h)^2} \;\;\text{(lower)}$$

Orange: upper semicircle. Blue: lower semicircle.

Example 2●●

Find the centre and radius of $x^2 + y^2 - 4x + 6y - 12 = 0$.

$x^2 - 4x + y^2 + 6y = 12$

Group the x terms and the y terms, and move the constant across.

$x^2 - 4x + 4 + y^2 + 6y + 9 = 12 + 4 + 9$

Complete the square twice. Add 4 and 9 to both sides.

$(x - 2)^2 + (y + 3)^2 = 25$

Centre $(2, -3)$, radius $5$.

Flashcards

Even powers y = xⁿ

U-shaped, symmetric about the y-axis, through $(-1, 1)$, $(0, 0)$, $(1, 1)$.

Odd powers y = xⁿ

Bottom-left to top-right, through $(-1, -1)$, $(0, 0)$, $(1, 1)$.

Single, double, triple zero

Crosses; touches and turns; flattens while crossing.

Circle equation

$(x - h)^2 + (y - k)^2 = r^2$

Domain and range of y = √x

$x \ge 0$, $y \ge 0$

Practice

1. Which graph passes through $(-1, 1)$?

   $y = x^3$$y = x^4$$y = x^5$$y = \sqrt{x}$

2. At $x = 2$, the graph of $y = (x - 2)^2(x + 1)$:

   crosses the axishas a horizontal point of inflectiontouches the axis and turnshas an asymptote

3. The domain of $y = \sqrt{9 - x^2}$ is:

   $-3 \le x \le 3$$0 \le x \le 3$$x \le 9$all real $x$

4. Find the radius of $x^2 + y^2 + 8y = 0$.

5. $y = x^3$ passes through:

   $(-1, 1)$$(1, -1)$$(0, 1)$$(-1, -1)$

6. A circle with centre $(0, 0)$ passes through $(3, 4)$. Find its radius.

7. How many different $x$-intercepts does $y = (x - 1)^2(x + 2)$ have?

   $1$$2$$3$$0$

8. The lower semicircle of radius $2$, centre the origin, is:

   $y = -\sqrt{4 - x^2}$$y = \sqrt{4 - x^2}$$y = -\sqrt{2 - x^2}$$x^2 + y^2 = 4$

Try again

More questions

New numbers every time. Keep going until you get five in a row.