# How do you find the vertex and intercepts for y=1/2(x-2)(x-4)?

Nov 3, 2017

x-ints at $x = 2 \text{ & } 4$, y-int at y=4.

Vertex at 3, -0.5

#### Explanation:

$x$ intercepts occur where $y = 0$. So you find them by setting y to 0, and solving.

Therefore if $\frac{1}{2} \left(x - 2\right) \left(x - 4\right) = y = 0$, the left hand side must be equal to zero. Since either $\left(x - 2\right)$ or $\left(x - 4\right)$ must equal zero, $\therefore x = 2 \text{ or } x = 4$.

Because parabolas are symmetrical, the vertex will occur half way between the $x$ intercepts. Halfway between 2 and 4 is 3. To find the y-value at this point, plug in $x = 3$ into the original equation:
$y = \frac{1}{2} \left(3 - 2\right) \left(3 - 4\right)$
$= \frac{1}{2} \cdot 1 \cdot - 1$
$= - \frac{1}{2}$
$\therefore$ vertex is at $\left(3 , - \frac{1}{2}\right)$

The $y$ intercept occurs when $x = 0$. So set x to 0 and solve:
$y = \frac{1}{2} \left(0 - 2\right) \left(0 - 4\right)$
$y = \frac{1}{2} \left(- 2\right) \left(- 4\right)$
$y = 4$

See graph for visual representation of these points:
graph{1/2(x-2)(x-4) [-5, 10, -3, 10]}