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mongoose mongoose
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11 years ago
I don't know how to find the vertex and the y-intercept when it is in factored form.
For example:
y=(x-4)(x+2)
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wrote...
#1
11 years ago
y = x^2 - 2x - 8
y = x^2 - 2x + ___ - 8 - ____
y = x^2 - 2x + 1 - 8 - 1
y = (x - 1)^2 - 9
vertex
(1, -9)

min/max (vertex) where absolutes of product factors equal
|x - 4| = |x + 2|
square both sides
x^2 - 8x + 16 = x^2 + 4x + 4
12 = 12x
x = 1
for x = 1, y = (-3)(3) = -9
(1, -9)
wrote...
#2
11 years ago
the graph would intersect the x-axis at x=4 and x=-2

The axis of symmetry would be half way bewteen them and passing though the vertex

half way is x=1

To find the y coordinate of the vertex let x=1

y=(x-4)(x+2)

y=(1-4)(1+2)

y=(-3)(3)

y=9

VERTEX........(1,9)

TO FIND THE Y-intercept let x=0 and find y

y=(x-4)(x+2)

y=(0-4)(0+2)

y=(-4)(2)

y=-8

y-intercept...................(0,-8)
wrote...
#3
11 years ago
Multiply the constants together to get -4 x 2 = -8

=> the y-intercept is (0,-8)

The curve crosses the y-axis at (-2,0) and (4,0)

The curve is symmetrical about the mid-point, i.e. x = 1

now, when x = 1, y = -9

so, the vertex is at (1, -9)..........which is also a minimum point of the function.

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