Expain how you could write a quadratic function in factored form that would have a vertex X coordinate of 3 and two distinct roots?

Two distinct roots means two real solutions for x (the parabola needs to cross the x-axis twice)

Vertex form of a quadratic equation: (h,k) is vertex

y = a(x-h)^2 + k

The x of the vertex needs to equal 3

y = a(x-3)^2 + k

In order to have two distinct roots the parabola must be (+a) upward facing with vertex below the x-axis or (-a) downward facing with vertex above the x-axis. Parabolas are symmetrical so for an easy factorable equation make "a" 1 or -1 depending on if you want the upward/downward facing one.

y = (x-3)^2 - 1

Vertex (3,-1) upwards facing with two distinct roots 4 and 2

y = x^2 -6x + 9 - 1

y = x^2 -6x + 8

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

Step-by-step explanation:

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this answer is !

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

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