The vertex of the graph Y 3 X-12 plus 2 would be -1/3 and -4/3. This is taught in math.
3x - 2y = 12 2y + 12 = 3x 2y = 3x - 12 y = 3x/2 - 6 Gradient = 3/2 Y-intercept = -6
y=-2x^2+8x+3
if ABCD is the triangle and O is the point then AO^2 + CO^2=BO^2+DO^2. Hence distance from 4th vertex can be calculated
let the vertex angle be x degrees, then the base angle is x + 9 degrees. Since in a triangle the sum of the angle is 180 degrees, and the base angles in an isosceles triangle are congruent, we have: x + 2(x + 9) = 180 x + 2x + 18 = 180 3x + 18 = 180 subtract 18 to both sides 3x = 162 divide by 3 to both sides x = 54 Thus the vertex angle is 54 degrees.
The vertex of the graph Y 3 X-12 plus 2 would be -1/3 and -4/3. This is taught in math.
Y=3x^2 and this is in standard form. The vertex form of a prabola is y= a(x-h)2+k The vertex is at (0,0) so we have y=a(x)^2 it goes throug (2,12) so 12=a(2^2)=4a and a=3. Now the parabola is y=3x^2. Check this: It has vertex at (0,0) and the point (2,12) is on the parabola since 12=3x2^2
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3x + 4y - 4 = 0 y = -2 3x + 4(-2) - 4 = 0 3x - 8 - 4 = 0 3x = 12 x = 4 (4,-2)
y = 2x2 + 3x + 6 Since a > 0 (a = 2, b = 3, c = 6) the graph opens upward. The coordinates of the vertex are (-b/2a, f(-b/2a)) = (- 0.75, 4.875). The equation of the axis of symmetry is x = -0.75.
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3x + 2y = 13Subtract 3x from each side:2y = -3x + 13Divide each side by 2:y = (-3/2)x + (13/2)The graph is a straight line, with a slope of (-3/2),through the point (0, 13/2) .
12 + 3x = 412 - 12 + 3x = 4 - 123x = -83x/3 = -8/3x = -8/3x = -(2 2/3)
The vertex coordinate point of the vertex of the parabola y = 24-6x-3x^2 when plotted on the Cartesian plane is at (-1, 27) which can also be found by completing the square.