Relations, Functions and Graphs II

Every quadratic graph is the same curve — a parabola — moved and stretched. Once you can find its turning point and where it crosses the axes, you can sketch any of them.

The shape of a quadratic

A quadratic function has the form

$$y = ax^2 + bx + c \qquad (a \ne 0)$$

The sign of $a$ decides the shape:

• $a > 0$ — the curve opens upwards, like a U, and has a minimum.
• $a < 0$ — the curve opens downwards, like an n, and has a maximum.

So $y = x^2 - 4x + 3$ is U-shaped, while $y = 5 + 4x - x^2$ is n-shaped, because the $x^2$ term is negative.

The $y$-intercept is always $c$, since putting $x = 0$ kills the other two terms. For $y = x^2 - 4x + 3$ the curve passes through $(0, 3)$.

Completing the square

Writing a quadratic as $a(x + p)^2 + q$ hands you the turning point directly.

Complete the square for $x^2 - 4x + 3$.

Halve the coefficient of $x$, then square it:

$$x^2 - 4x = (x-2)^2 - 4$$

The $-4$ appears because $(x-2)^2$ expands to $x^2 - 4x + 4$, so 4 must be taken back off. Now bring the $+3$ along:

$$x^2 - 4x + 3 = (x-2)^2 - 4 + 3 = (x-2)^2 - 1$$

Check by expanding: $x^2 - 4x + 4 - 1 = x^2 - 4x + 3$ ✓

Another: $x^2 + 6x + 5$.

Half of 6 is 3, so

$$x^2 + 6x + 5 = (x+3)^2 - 9 + 5 = (x+3)^2 - 4$$

When the $x^2$ term is negative

Take the minus out first.

$$5 + 4x - x^2 = -(x^2 - 4x) + 5 = -\left[(x-2)^2 - 4\right] + 5 = -(x-2)^2 + 4 + 5 = -(x-2)^2 + 9$$

Check at $x = 2$: the original gives $5 + 8 - 4 = 9$ ✓

Maximum and minimum points

From the completed square $y = a(x+p)^2 + q$, the turning point is at $(-p, q)$.

A square is never negative, so:

• If $a > 0$, the smallest $y$ can be is $q$, reached when the bracket is zero. That is a minimum.
• If $a < 0$, the largest $y$ can be is $q$. That is a maximum.

$y = (x-2)^2 - 1$ has a minimum at $(2, -1)$, because $(x-2)^2$ is smallest when $x = 2$.

$y = -(x-2)^2 + 9$ has a maximum at $(2, 9)$.

The axis of symmetry is the vertical line through the turning point, so it is $x = -p$: for $y = x^2 - 4x + 3$ that is $x = 2$. The curve is a mirror image on either side of it.

Roots and intercepts

The roots are the $x$ values where $y = 0$ — where the curve crosses the $x$-axis.

Find the roots of $y = x^2 - 4x + 3$.

$$x^2 - 4x + 3 = 0$$ $$(x-1)(x-3) = 0$$ $$x = 1 \quad \text{or} \quad x = 3$$

The curve crosses the $x$-axis at $(1,0)$ and $(3,0)$.

How many roots?

The turning point tells you at a glance:

So $y = x^2 - 6x + 11 = (x-3)^2 + 2$ never crosses the $x$-axis: its lowest point is 2 units above it.

Sketching a quadratic

Four pieces of information make a sketch worth full marks:

1. The shape — from the sign of $a$.
2. The $y$-intercept — the value of $c$.
3. The roots, if there are any — by factorising.
4. The turning point — by completing the square, or from the midpoint of the roots.

For $y = x^2 - 4x + 3$: U-shaped; through $(0,3)$; roots at $x = 1$ and $x = 3$; minimum at $(2,-1)$.

That is enough to draw the curve accurately and to label everything the question asks for.

Solving with a line

Where a curve and a line meet, both equations hold, so set them equal.

Find where $y = x^2$ meets $y = x + 2$.

$$x^2 = x + 2$$ $$x^2 - x - 2 = 0$$ $$(x-2)(x+1) = 0$$ $$x = 2 \quad \text{or} \quad x = -1$$

Substitute back into the simpler equation, $y = x + 2$:

• $x = 2$ gives $y = 4$
• $x = -1$ gives $y = 1$

They meet at $(2, 4)$ and $(-1, 1)$.

Check in $y = x^2$: $2^2 = 4$ ✓ and $(-1)^2 = 1$ ✓