$7(x-3)^2-4(x-3)-3\\\\substitute\ t=x-3,\ then\ we\ have\\\\7t^2-4t-3\\\\a=7;\ b=-4;\ c=-3\\\\\Delta=b^2-4ac\\\\\Delta=(-4)^2-4\cdot7\cdot(-3)=16+84=100\\\\t_1=\dfrac{-b-\sqrt\Delta}{2a};\ t_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{100}=10\\\\t_1=\dfrac{-(-4)-10}{2\cdot7}=\dfrac{4-10}{14}=\dfrac{-6}{14}=-\dfrac{3}{7}\\\\t_2=\dfrac{-(-4)+10}{2\cdot7}=\dfrac{4+10}{14}=\dfrac{14}{14}=1\\\\7t^2-4t-3=7\left(t+\dfrac{3}{7}\right)(t-1)=(7t+3)(t-1)\\\\=(7(x-3)+3)(x-3-1)=(7x-21+3)(x-4)\\\\=\boxed{(7x-18)(x-4)}$

7 0

3 years ago

What are the coordinates of the image of R for a dilation with center (0,0) and scale factor 3?

OverLord2011 [107]

Answer:

(3, -3)

Step-by-step explanation:

If a point is dilated about the origin rule to be followed,

(x, y) → (kx, ky)

where k is a scale factor by which the point is being dilated.

If a point R having coordinates (1, -1) is dilated by scale factor of 3, coordinates of the image will be [3×1, 3(-1)]

Therefore, coordinates of the image R' will be (3, -3)

5 0

3 years ago