by plotting the graph of parabola y=2+3x-2x^2 which 3 option are true

1) the graph is the same as that of y=6+9x-6x^2
2)the parabola has a minimum point
3)the gradient of the parabola at x=2 is 0
4)the graphy=2+3x-2x^2crosses the x-axis at the same point as the graph y=6+9x-6x^2
5)the vertex of the parabola is at(1,5)
6)the equation of the axis of symmetery is y=-1
7)the gradient of the parabola is positive when x=0
8)the intercept on the y axis is at y=2

Have you plotted this on your calculator?

To determine which options are true, let's analyze each statement one by one.

1) The graph is the same as that of y=6+9x-6x^2.
To determine if two graphs are the same, we can compare the equations. The given equation, y=2+3x-2x^2, is not the same as y=6+9x-6x^2 because the constants in both equations are different. Therefore, this statement is FALSE.

2) The parabola has a minimum point.
To determine if a parabola has a minimum or maximum point, we can look at the coefficient of x^2. In this case, the coefficient is negative (-2x^2), which means the parabola opens downwards and has a maximum point. Therefore, this statement is FALSE.

3) The gradient of the parabola at x=2 is 0.
The gradient of a function at a specific point can be found by taking the derivative of the function and evaluating it at that point. Taking the derivative of y=2+3x-2x^2, we get dy/dx = 3-4x. Evaluating this at x=2, we have dy/dx = 3-4(2) = 3-8 = -5. Since the gradient is -5, this statement is FALSE.

4) The graph y=2+3x-2x^2 crosses the x-axis at the same point as the graph y=6+9x-6x^2.
To find where a function crosses the x-axis, we can set y=0 and solve for x. For y=2+3x-2x^2, setting y=0 gives us 2+3x-2x^2=0. Similarly, for y=6+9x-6x^2, we have 6+9x-6x^2=0. Solving both equations may give us different x-values, so this statement is FALSE.

5) The vertex of the parabola is at (1,5).
To find the vertex of a parabola in the form y=ax^2+bx+c, we can use the formula x=-b/(2a) to find the x-coordinate of the vertex. In this case, a=-2, b=3, and c=2. Plugging these values into the formula, we get x=-3/(2*(-2)) = -3/(-4) = 3/4. To find the y-coordinate, we substitute this value of x into the equation: y=2+3(3/4)-2(3/4)^2 = 2+9/4-9/8 = 16/8+18/8-9/8 = 25/8 = 3.125. Therefore, the vertex is not at (1,5), so this statement is FALSE.

6) The equation of the axis of symmetry is y=-1.
The equation of the axis of symmetry for a parabola in the form y=ax^2+bx+c is x=-b/(2a). In this case, a=-2, and b=3. Plugging these values into the formula, we get x=-3/(2*(-2)) = -3/(-4) = 3/4. Therefore, the equation of the axis of symmetry is x=3/4, not y=-1. This statement is FALSE.

7) The gradient of the parabola is positive when x=0.
To find the gradient (slope) of the parabola at a specific point, we can evaluate the derivative of the function at that point. Taking the derivative of y=2+3x-2x^2, we get dy/dx = 3-4x. Evaluating this at x=0, we have dy/dx = 3-4(0) = 3. Since the gradient is 3, which is positive, this statement is TRUE.

8) The intercept on the y-axis is at y=2.
The y-intercept is the point where the parabola crosses the y-axis, meaning x=0. Substituting x=0 into y=2+3x-2x^2, we have y=2+3(0)-2(0)^2 = 2. Therefore, this statement is TRUE.

Based on the analysis above, the true statements are:
7) The gradient of the parabola is positive when x=0.
8) The intercept on the y-axis is at y=2.