Answer
a) $\left(-\dfrac{b}{2a},-\dfrac{b^2-4ac}{4a}\right)$
b) $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ for $b^2-4ac>0$
Work Step by Step
We are given the function:
$f(x)=ax^2+bx+c,$ where $a\not=0$
a) Rewrite the function by completing the square:
$f(x)=a\left(x^2+\dfrac{b}{a}x\right)+c$
$=a\left(x^2+2\cdot\dfrac{b}{2a}x+\dfrac{b^2}{4a^2}-\dfrac{b^2}{4a^2}\right)+c$
$=a\left(x^2+2\cdot\dfrac{b}{2a}x+\dfrac{b^2}{4a^2}\right)-a\cdot \dfrac{b^2}{4a^2}+c$
$=a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a}$
The function $f$ reaches its minimum or maximum depending on the term $a\left(x+\dfrac{b}{2a}\right)^2$. The extremum is obtained for $a\left(x+\dfrac{b}{2a}\right)^2=0$. This means:
$x+\dfrac{b}{2a}=0\Rightarrow x=-\dfrac{b}{2a}$
Determine the corresponding value of $f$:
$f\left(-\dfrac{b}{2a}\right)=a\left(-\dfrac{b}{2a}+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a}$
$=0-\dfrac{b^2-4ac}{4a}$
$=-\dfrac{b^2-4ac}{4a}$
The coordinates of the vertex are: $\left(-\dfrac{b}{2a},-\dfrac{b^2-4ac}{4a}\right)$.
b) The zeroes of the function are the solutions of the equation:
$f(x)=0$
$a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a}=0$
The equation crosses the $x$-axis twice when it has two different roots. We have:
$\left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2-4ac}{4a^2}$
The equation has two solutions if the right side is greater than zero, which measn $b^2-4ac>0$.
$x+\dfrac{b}{2a}=\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$
$x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
