Today's lesson
Not every quadratic factorises neatly. The quadratic formula finds the $x$-intercepts of any parabola, even when they are irrational. The discriminant tells us how many $x$-intercepts there are before we even solve — and the axis-of-symmetry rule $x=-\tfrac{b}{2a}$ locates the turning point.
Learning intentions
- Use the quadratic formula to find the $x$-intercepts of $y=ax^2+bx+c$
- Use the discriminant $\Delta=b^2-4ac$ to find the number of $x$-intercepts
- Use $x=-\tfrac{b}{2a}$ (the axis of symmetry) to locate the turning point
- Sketch the parabola, labelling all significant points
Part 1 — The formula and the discriminant (~8 min)
To sketch $y=ax^2+bx+c$: the $y$-intercept is $(0,c)$; the $x$-intercepts come from solving $ax^2+bx+c=0$ with the quadratic formula; and the turning point sits on the axis of symmetry $x=-\tfrac{b}{2a}$.
Key ideas
- Quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.
- Discriminant $\Delta=b^2-4ac$ controls the number of $x$-intercepts:
$\Delta<0\Rightarrow$ no $x$-intercepts; $\Delta=0\Rightarrow$ one; $\Delta>0\Rightarrow$ two. - Axis of symmetry / turning point: $x=-\tfrac{b}{2a}$; substitute back to get the $y$-coordinate.
Part 2 — Discriminant & turning point (Example 1, ~10 min)
- a $\Delta=(-6)^2-4(3)(5)=36-60=-24<0$, so there are no $x$-intercepts.
- b $y$-intercept is $(0,5)$.
- c $x=-\tfrac{b}{2a}=-\tfrac{-6}{2(3)}=1$; then $y=3(1)^2-6(1)+5=2$. Turning point $(1,2)$, a minimum (the whole curve sits above the $x$-axis).
Now you try: For $y=2x^2-4x+1$ find the number of $x$-intercepts, the $y$-intercept and the turning point. Answers: $\Delta=8>0$ so two $x$-intercepts; $y$-intercept $(0,1)$; turning point $(1,-1)$ (a minimum).
Part 3 — Sketching with the formula (Example 2, ~14 min)
$y$-intercept: $(0,-3)$.
$x$-intercepts ($y=0$): $2x^2+4x-3=0$, so $x=\dfrac{-4\pm\sqrt{4^2-4(2)(-3)}}{2(2)}=\dfrac{-4\pm\sqrt{40}}{4}=0.58,\ -2.58$ (to 2 d.p.). Intercepts $(0.58,0)$ and $(-2.58,0)$.
Turning point: $x=-\tfrac{b}{2a}=-\tfrac{4}{2(2)}=-1$; then $y=2(-1)^2+4(-1)-3=-5$, so $(-1,-5)$ (a minimum).
1. $y$-intercept = $(0,c)$.
2. $\Delta=b^2-4ac$ → how many $x$-intercepts.
3. $x$-intercepts from $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.
4. Turning point at $x=-\dfrac{b}{2a}$, then substitute for $y$.
5. Join with a smooth curve.
Now you try: Sketch $y=3x^2-6x+1$. Answers: $y$-intercept $(0,1)$; $x$-intercepts $\approx0.18$ and $1.82$; turning point $(1,-2)$ (a minimum).
📺 Walkthrough: using the discriminant and the quadratic formula on $y=2x^2+4x-3$, then locating the turning point with $x=-\tfrac{b}{2a}$.
Practice 3.1 — state the number of $x$-intercepts using the discriminant.
- $y=x^2-4x+4$
- $y=x^2+2x+5$
- $y=x^2-3x-1$
- $y=2x^2-3x+5$
Part 4 — Quick quiz (5 min)
Pick the correct answer for each, then click Mark.
Q1. The discriminant of $y=ax^2+bx+c$ is:
Q2. If $\Delta<0$, the parabola has:
Q3. For $y=2x^2+4x-3$, the axis of symmetry $x=-\tfrac{b}{2a}$ is:
Q4. The $y$-intercept of $y=3x^2-6x+5$ is:
Q5. $x^2-2x+3=0$ has $\Delta=4-12=-8$. The number of $x$-intercepts is:
Working program — Cambridge Ex 7E (p622)
After the quiz, open Cambridge Chapter 7 (page 622) and complete the following:
| Set work | Extension |
|---|---|
| Questions 1–6 (½), 8, 9 | Questions 10–12 |
"(½)" means do every second part. Show the discriminant and the quadratic-formula working for any sketch.
Exit ticket — write in your book
- What does the discriminant $b^2-4ac$ tell you about a parabola?
- Write down the quadratic formula from memory.
- How do you find the turning point once you know $a$, $b$ and $c$?