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).
y = 2x2 - 3x + 1 y = (x - 1)(2x - 1) x - 1 = 0 2x - 1 = 0 x = 1 x = 1/2 the answer is 2
The tangent to the curve y = 2x2 - 3x + 2 at x=3 first we differentiate the eqution and get dy/dx = 4x -3 feed in the x=3 dy/dx = 9 So we now know that the gradient of the tangent is 9 We can also find out one of the points that is passes through. y = 2x2 - 3x + 2 when x = 3 y = 2*32 - 3*3 + 2 y = 29 So we have a line of gradient 9 that passes through the pint (3,29) We can now find out the y intercept 29 - (3*9) = 2 y = 3x + 2
-5
3x - y = 12 is the same as 3x - 12 = y, or y = 3x - 12.
2x2 + 3x - 20 = (x + 4)(2x - 5).
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).
No.
Solve by factoring.2x2 - 3x + 1 = 0
y=2x2+3x+1
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 - 3x + 1 y = (x - 1)(2x - 1) x - 1 = 0 2x - 1 = 0 x = 1 x = 1/2 the answer is 2
If you graph this it will be a parabola that has each end pointing down. It crosses the x-axis at: y=-2x^2+3x+5 y=-1(2x^2-3x-5) y=-1(2x-5)(x+1) y=5/2 y=-1
3x-5 = y2x^2 + (3x+5)^2 = 12911x^2 +30x + 25 = 12911x^2 + 30x - 104 = 0solve using quadratic formula x = +2y = 3x-5 = 1Another Answer:-If: 3x-y = 5 and 2x2+y2 = 129Then: y = 3x-5 and so 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0Solving the quadratic equation: x = -2 and x = 52/11The solutions by substitution are: x = -2, y = -11 and x = 52/11, y = 101/11
15x - 3xy = 3x(5 - y) 20 - 4y = 4(5 - y) 15x - 3xy + 20 - 4y = (3x + 4)(5 - y)
It solves a system of equations. Think of the 2X2 case of having X+Y=1 and 3X+2Y=6.
Subtracting the second equation from first: 3x - y = 25 3x + 4y = 5 0x - 5y = 20 5y = -20 y = -4 Plugging y = -4 into either equation gives us x = 7, So x = 7 and y = -4