Quadratic functions practice questions — IB Maths AA

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

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: $$r^2+s^2=7^2-2(10)=49-20=29$$

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: $$2<x<3$$

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

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

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

Solution: $(2,3)$.

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

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

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

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

Factoring gives $(x-2)(x-3)=0$, so the roots are $2$ and $3$. Their sum is $5$. (Equivalently, $-b\mathrm{/}a=5\mathrm{/}1$.)