Quadratic Functions Worked ExamplesIGCSE Maths 0580

(a) The graph crosses the $x$-axis where $y = 0$. Factorise.

(b) The $y$-intercept is the constant term, found by setting $x = 0$.

(c) The line of symmetry sits exactly halfway between the two roots.

(d) Substitute $x = 2$ to find the $y$-coordinate of the turning point.

Halve the coefficient of $x$ and square it.

$(x-3)^2$ expands to $x^2 - 6x + 9$, which is $2$ short of the constant we need.

Simplify.

A square is never negative, so the smallest value is $2$, reached when $x = 3$.

Factor the leading coefficient out of the first two terms only.

Complete the square inside the bracket.

Substitute back — the $-4$ is inside the bracket, so it gets multiplied by $2$.

Simplify.

Match with $p(x + q)^2 + r$. The bracket is $(x - 2)$, so $q = -2$.

(b) $2(x - 2)^2$ is never negative, so the expression is smallest when that term is $0$, at $x = 2$.

(a) Set $y = 0$. Multiplying through by $-1$ makes factorising easier.

Read off the roots.

(b) The line of symmetry is midway between the roots.

Substitute to find $y$.

The $x^2$ coefficient is negative, so the curve is an inverted U and the turning point is a maximum.

Equal roots means the discriminant is zero — the curve touches the $x$-axis rather than crossing it.

Substitute $a = 1$, $b = k$, $c = 9$.

Solve.

Check $k = 6$: $x^2 + 6x + 9 = (x+3)^2$, a perfect square with the repeated root $x = -3$. ✓

Two sides of length $x$ use $2x$ metres, leaving the rest for the side parallel to the wall.

Area is length times width.

This is a negative quadratic, so it has a maximum. Find the roots to locate the line of symmetry.

The maximum lies midway between the roots.

Find the greatest area.

(a) The highest point of the curve is on grid lines, so it can be read exactly.

(b) The curve $y = a(x + p)^2 + q$ has its turning point at $(-p, q)$. Match this to $(-1, 4)$.

The turning point does not fix $a$, so use one more point on the curve. The $y$-intercept $(0, 2)$ is easiest.

Check with a second grid point on the curve, $(1, -4)$.

(c) Set $f(x) = 0$ and isolate the bracket.

Square root both sides, remembering both signs.

Since $\sqrt{2} \approx 1.41$, the roots are about $-2.41$ and $0.41$. The curve crosses the $x$-axis at about $-2.4$ and $0.4$, so the answers are sensible. ✓

(a) Read the highest point of the curve. It is one small square above the $8$ line, level with $t = 1.2$.

(b) Draw the horizontal line $h = 6$. It cuts the curve twice, at about $t = 0.54$ and $t = 1.86$. The ball is above $6$ m between these times.

(c) Expand the given form and simplify.

$5(t - 1.2)^2$ is never negative, so $h$ is greatest when that term is $0$, at $t = 1.2$, giving $h = 8.2$. This confirms part (a).

(d) Rearrange into the form $at^2 + bt + c = 0$.

Use the quadratic formula with $a = 5$, $b = -12$, $c = 5$.

Subtract the unrounded values.

(a) The arch is symmetrical, so its highest point is halfway between the feet $(0, 0)$ and $(8, 0)$, at $(4, 6)$. Start from turning-point form.

The arch passes through $O(0, 0)$. Substitute to find $a$.

Expand to reach the given form.

(b) The lorry is centred on $x = 4$, so its top corners are $1.6$ m either side, at $x = 2.4$ and $x = 5.6$. By symmetry the arch has the same height at both, so check one.

The arch is $5.04$ m high above the edges of the lorry. This is more than $5$ m, so the lorry fits, with $4$ cm to spare.

(c) Let the load reach a distance $d$ m either side of the centre line. The arch must be $4.5$ m high there.

So $d = 2$. This is the half-width, so double it.