Quadratic Function practice questions — A-Level Maths

$\Delta = 36 - 32 = 4$, $a = 1$, so $f_{\min} = -\dfrac{4}{4} = -1$ (attained at $x = 3$).

Factor the quadratic: $$x^2 - 5x + 6 = (x-2)(x-3)$$ The roots are at $x = 2$ and $x = 3$. The leading coefficient is positive, so the parabola opens upward and the expression is negative between the roots: $$2 < x < 3$$

$x^{2} - 5x + 6 = (x - 2)(x - 3)$.

The product $(x-2)(x-3) < 0$ when exactly one factor is negative — i.e. when $2 < x < 3$.

Solution: $(2, 3)$.

$\Delta = (m - 2)^2 - 4 = 0 \Rightarrow (m - 2)^2 = 4 \Rightarrow m - 2 = \pm 2 \Rightarrow m \in \{0, 4\}$.

Factorise the quadratic: $$x^2 - 2x - 8 = (x-4)(x+2)$$ The curve crosses the $x$-axis at: $$x = -2 \quad\text{and}\quad x = 4$$ The parabola opens upward, so $y \leqslant 0$ between the roots: $$-2 \leqslant x \leqslant 4$$

$(x - 2)(x + 2) < 0 \Leftrightarrow x \in (-2, 2)$.

By Vieta's formulas, the sum of the roots is $-\frac{b}{a}$ and the product is $\frac{c}{a}$: $$r+s=7$$ $$rs=10$$ Use the identity that rewrites $r^2+s^2$ in terms of the sum and product: $$r^2+s^2=(r+s)^2-2rs$$ Substitute the known values: $$r^2+s^2=7^2-2(10)=49-20=29$$

$x_1 + x_2 = -\dfrac{-7}{2} = \dfrac{7}{2}$. (And the product is $\dfrac{c}{a} = \dfrac{3}{2}$.)

$f(2) = 12 - 14 + 2 = 0$. So $f(2025) \cdot f(2) = f(2025) \cdot 0 = 0$, regardless of $f(2025)$.

$(x - 2)(x - 3) = 0 \Rightarrow x \in \{2, 3\}$.