Coaching Centre
Quadratic models
The parabola y = ax² + bx + c, its intercepts, axis of symmetry and vertex found from the midpoint of the x-intercepts, and quadratic models of projectiles, areas and profit with a sensible domain.
Throw a ball, aim a garden hose or look at a bridge arch, and you see the same curve: a parabola. Its equation has an $x^2$ term, and its key feature is a turning point, the highest or lowest value. In practical problems that turning point is almost always the answer someone wants: the greatest height, the largest area, the most profit.
The shape of y = ax² + bx + c
- If $a \gt 0$ the parabola opens upward and the vertex is a minimum.
- If $a \lt 0$ it opens downward and the vertex is a maximum.
- The $y$-intercept is $c$, because putting $x = 0$ leaves only $c$.
- The curve is symmetric about a vertical line, the axis of symmetry, which passes through the vertex.
If the $x$-intercepts are $x = p$ and $x = q$, the axis of symmetry is halfway between them:
$x = \dfrac{p + q}{2}$
Substitute this $x$ into the equation to get the $y$-value of the vertex.
Why the midpoint? The parabola is a mirror image of itself across the axis, so the two points where it crosses the $x$-axis are the same distance from the axis on either side.
For $y = x^2 - 6x + 5$, find the intercepts, the axis of symmetry and the vertex.
On Kicked ball, check that the axis of symmetry is midway between $0$ and $40$, and read the rule. Now move the right intercept to $50$ while keeping the height at $10$: what happens to the number in front of $x^2$? On Fountain jet, find the $y$-intercept and explain why the model only makes sense between the two intercepts.
Projectiles
A football's flight is modelled by $h = -0.025x^2 + x$, where $x$ m is the horizontal distance and $h$ m the height. Find where it lands, its greatest height, and its height $10$ m from the kicker.
Area and profit
A farmer has $40$ m of fencing to make a rectangular pen against a long shed wall, so only three sides need fencing. If each side touching the wall is $x$ m, the area is $A = x(40 - 2x)$. Find the largest area.
A stall's weekly profit when it charges $\$x$ per item is $P = -2x^2 + 80x - 600$. A graphing app shows $P = 0$ at $x = 10$ and $x = 30$. What price gives the most profit, and what is that profit?
Your turn
A water jet follows $h = -0.25x^2 + 2.25$, with $x$ and $h$ in metres. It hits the ground at $x = -3$ and $x = 3$. Find its greatest height and where it is.
Show the working
The axis is $x = \dfrac{-3 + 3}{2} = 0$. Then $h = -0.25 \times 0 + 2.25 = 2.25$ m. The greatest height is $2.25$ m, directly above the nozzle at $x = 0$.
Which parabola opens downward?
Only $-x^2 + 4x$ has a negative coefficient of $x^2$ ($a = -1$).
What is the $y$-intercept of $y = x^2 - 6x + 5$?
Put $x = 0$: $y = 0 - 0 + 5 = 5$.
A parabola cuts the $x$-axis at $x = 1$ and $x = 5$. What is the axis of symmetry? Give the $x$-value.
$x = (1 + 5) \div 2 = 3$.
Find the $y$-value of the vertex of $y = x^2 - 6x + 5$.
The axis is $x = 3$. $y = 9 - 18 + 5 = -4$.
A ball follows $h = -0.025x^2 + x$. How far from the kicker does it land, in metres?
$x(-0.025x + 1) = 0$ gives $x = 0$ or $x = 1 \div 0.025 = 40$.
For the same ball, what is the greatest height, in metres?
The axis is $x = 20$. $h = -0.025 \times 400 + 20 = -10 + 20 = 10$.
For the same ball, what is the height $30$ m from the kicker?
$h = -0.025 \times 900 + 30 = -22.5 + 30 = 7.5$ m. It matches the height at $10$ m because of the symmetry about $x = 20$.
For the pen with area $A = x(40 - 2x)$, which domain is sensible?
The side must be positive, and $40 - 2x$ must also be positive, so $x \lt 20$.
Exam style: Weekly profit is $P = -2x^2 + 80x - 600$ for a price of $\$x$, and $P = 0$ when $x = 10$ and $x = 30$. What price, in dollars, gives the greatest profit?
The axis of symmetry is midway between $10$ and $30$: $x = (10 + 30) \div 2 = 20$.
Exam style: What is that greatest weekly profit, in dollars?
$P = -2 \times 20^2 + 80 \times 20 - 600 = -800 + 1600 - 600 = 200$.