Quadratics

The same parabola can be written three ways. Each one answers a different question at a glance, and the SAT often hands you the form that makes the question quick.

Form	Shows
$y = a (x - r) (x - s)$	Zeros $r, s$
$y = a (x - h)^{2} + k$	Vertex $(h, k)$
$y = a x^{2} + b x + c$	$y$ -intercept $c$

The vertex's

x

- \frac{b}{2 a}

, halfway between the zeros. Sum of solutions is

- \frac{b}{a}

, and the product is

\frac{c}{a}

Work through it

$y = (x - 2) (x - 8)$ . Find the minimum value.

Halfway between the zeros. $x = 5$

Substitute. $(3) (- 3) = - 9$

What's the sum of the solutions of $2 x^{2} - 10 x + 3 = 0$ ?

No need to solve. $- \frac{b}{a} = \frac{10}{2} = 5$

For what $c$ does $x^{2} + 6 x + c = 0$ have one solution?

Discriminant zero. $36 - 4 c = 0$

Solve. $c = 9$

Watch out for this

Reading $(x - 3)^{2}$ 's vertex as $x = - 3$ .

$x - 3 = 0$ at $x = 3$ . The sign flips.

The short version
Factored form shows the zeros, vertex form shows the vertex, and

b^{2} - 4 a c

tells you how many real solutions there are.

Your turn

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