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Title: How do you find the vertex of a quadratic function in factored form? Post by: mongoose on Nov 7, 2012 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) Title: How do you find the vertex of a quadratic function in factored form? Post by: money2011 on Nov 7, 2012 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) Title: How do you find the vertex of a quadratic function in factored form? Post by: moneyboy396 on Nov 7, 2012 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) Title: How do you find the vertex of a quadratic function in factored form? Post by: _biology on Nov 7, 2012 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. :)> |