How do you find the vertex and intercepts for y=1/2(x-2)(x-4)y=12(x−2)(x−4)?

1 Answer
TreeReader
Nov 3, 2017

x-ints at x = 2" & "4x=2 & 4, y-int at y=4.

Vertex at 3, -0.5

Explanation:

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

Therefore if 1/2(x-2)(x-4)=y=012(x−2)(x−4)=y=0, the left hand side must be equal to zero. Since either (x-2)(x−2) or (x-4)(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]}

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