Question

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

Answer 1

x-ints at `x = 2" & "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 `1/2(x-2)(x-4)=y=0`, the left hand side must be equal to zero. Since either `(x-2)` or `(x-4)` must equal zero, `:. x=2 " 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=1/2(3-2)(3-4)`
`=1/2*1*-1`
`=-1/2`
`:.` vertex is at `(3,-1/2)`

The `y` intercept occurs when `x=0`. So set x to 0 and solve:
`y=1/2(0-2)(0-4)`
`y=1/2(-2)(-4)`
`y=4`

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