F(x) = -x^2 - x +2
graph function identify the vertex and axis of symmerty.
Can some one explain the steps to get answer? I know I need to find the points on a graph just so confused
learly you are studying the quadratic function , which when graphed is a parabola.
One of the first things you learn is how to find the vertex. Go to your text or you class notes to look at the example and do the same thing.
f(x) = -(x^2+x-2)
= -(x+2)(x-1)
The roots are at -2 and +1
So, the vertex and the axis of symmetry are midway between the roots, at x = -1/2
Now just find the y-coordinate of the vertex.
To find the vertex and axis of symmetry of the function f(x) = -x^2 - x + 2, you need to follow these steps:
1. Identify the coefficient of x^2: In this case, the coefficient is -1.
2. Find the x-coordinate of the vertex: The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a is the coefficient of x^2 and b is the coefficient of x.
In our case, a = -1 and b = -1, so plugging these values into the formula gives x = -(-1) / (2 * -1) = 1/2.
Therefore, the x-coordinate of the vertex is x = 1/2.
3. Substitute the x-coordinate into the function to find the y-coordinate of the vertex: Replace the x in the function with the x-coordinate of the vertex and calculate the corresponding y-coordinate.
Substituting x = 1/2 into f(x), we get f(1/2) = -(1/2)^2 - (1/2) + 2 = -1/4 - 1/2 + 2 = 7/4.
Thus, the y-coordinate of the vertex is y = 7/4.
4. Determine the axis of symmetry: The axis of symmetry is a vertical line that passes through the vertex. Since the x-coordinate of the vertex is 1/2, the equation of the axis of symmetry is x = 1/2.
Finally, to graph the function, plot the vertex, which is located at (1/2, 7/4), and draw a parabola that opens downwards. The axis of symmetry is the vertical line x = 1/2. You can also plot a few more points on the graph to see the shape of the parabola.