Quadratic Formula & Discriminant (Ex 7E)
Year 10 Mathematics Core · Cambridge Ch 7, §7E · Mr Wong
Learning intentions.
- Use the quadratic formula to find $x$-intercepts (including irrational ones)
- Use the discriminant $\Delta=b^2-4ac$ to find the number of $x$-intercepts
- Use $x=-\tfrac{b}{2a}$ to locate the turning point, then sketch
Key results
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.
$\Delta=b^2-4ac$: $\Delta<0$ → no $x$-int, $\Delta=0$ → one, $\Delta>0$ → two.
Turning point at $x=-\tfrac{b}{2a}$ (then substitute for $y$). $y$-intercept $(0,c)$.
Warm-up — the discriminant
1. Find the discriminant $\Delta=b^2-4ac$ and state the number of $x$-intercepts.
a $y=x^2-2x-3$ $\Delta=$
b $y=x^2+6x+9$ $\Delta=$
c $y=x^2+x+4$ $\Delta=$
d $y=2x^2-5x+1$ $\Delta=$
2. Use the quadratic formula to find the $x$-intercepts (exact form).
a $y=x^2-2x-5$
b $y=x^2+4x+2$
Part A — Turning point with $x=-\tfrac{b}{2a}$
3. Find the turning point of each (give $x=-\tfrac{b}{2a}$, then the $y$-value).
a $y=x^2-6x+11$
b $y=2x^2+4x-3$
Part B — Sketching
4. Sketch $y=x^2-2x-3$: find the $y$-intercept, the $x$-intercepts and the turning point, then draw the curve.
a $y$-intercept
b $x$-intercepts
c Turning point
Parabolas 7E — continued
Part B (sketching) · Part C (irrational intercepts) · Challenge
Part B — Sketching (continued)
5. Sketch $y=2x^2+4x-3$ (round $x$-intercepts to 2 d.p.): find the $y$-intercept, the $x$-intercepts and the turning point.
a $y$-intercept
b $x$-intercepts
c Turning point
Part C — Irrational intercepts
6. Use the quadratic formula to find the $x$-intercepts to 2 d.p. State "none" if there are none.
a $y=x^2-4x+1$
b $y=x^2+3x+5$
c $y=2x^2-6x+1$
d $y=3x^2-6x+1$
Challenge
7. For what values of $k$ does $y=x^2+kx+9$ have exactly
one $x$-intercept?
(Hint: set the discriminant to zero.)