Q: What are the points of y equals 2x-2?

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no

No.

If: 3x-y = 5 and 2x2+y2 = 129 Then: 3x-y = 5 => y = 3x-5 And so: 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0 Using the quadratic equation formula: x = 52/11 and x = -2 By substitution points of intersection are: (52/11, 101/11) and (-2, -11)

2x2 + 3x - 20 = (x + 4)(2x - 5).

The locus of points equidistant from lines y = 0 and x = 3 is the line y = -x + 3.

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no

No.

y = 2x2 - 3x - 20 ∴ y = 2x2 - 8x + 5x - 20 ∴ y = 2x(x - 4) + 5(x - 4) ∴ y = (2x + 5)(x - 4) The x intercepts occur when y = 0, which happens when either of the two factors on the left side of the equation equal zero. This means that they are at the points (-5/2, 0) and (4, 0).

If: 3x-y = 5 and 2x2+y2 = 129 Then: 3x-y = 5 => y = 3x-5 And so: 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0 Using the quadratic equation formula: x = 52/11 and x = -2 By substitution points of intersection are: (52/11, 101/11) and (-2, -11)

y=2x2+5. Plug in 0 for x, and you get y=2(0)^2 +5 =0+5=5. 5 is the minimum value for y.

2x2 + 3x - 20 = (x + 4)(2x - 5).

Assuming the missing symbol there is an equals sign, then we have: y - 2x2 - 4x = 4 We can find it's vertex very easily by solving for y, and finding where it's derivative equals zero: y = 2x2 + 4x + 4 y' = 4x + 4 0 = 4x + 4 x = -1 So the vertex occurs Where x = -1. Now we can plug that back into the original equation to find y: y = 2x2 + 4x + 4 y = 2 - 4 + 4 y = 2 So the vertex is at the point (-1, 2)

Y Equals X PointsAll points that has the same y coordinates as x coordinates are on the y=x line.

-5

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1.249 radians from the horizontal.