Find the vertex and the axis of symmetry, and graph the function.
f(x) = x^2 - 5x
Graphing is not my strong point, I really need help with this. Thanks
there are oodles of online graphing utilities.
wolframalpha.com
desmos.com
come to mind. Just type in your function.
To find the vertex and axis of symmetry of the function f(x) = x^2 - 5x, we can use the formula:
The vertex of the parabola with equation f(x) = ax^2 + bx + c is given by point V(xv, yv) where:
xv = -b/2a
yv = f(xv)
For our given function f(x) = x^2 - 5x, we can identify that a = 1, b = -5, and c = 0. Now, let's calculate the vertex:
xv = -(-5) / (2 * 1) = 5/2 = 2.5
To find yv, substitute xv into the function:
yv = f(2.5) = (2.5)^2 - 5(2.5)
= 6.25 - 12.5
= -6.25
So, the vertex is V(2.5,-6.25).
To find the axis of symmetry, we use the x-coordinate of the vertex, which is 2.5. Therefore, the axis of symmetry is the vertical line x = 2.5.
Now let's graph the function:
To graph f(x) = x^2 - 5x, we start by plotting the vertex (2.5,-6.25) on the coordinate plane. Since the coefficient of x^2 is positive, the parabola opens upwards.
Next, choose some values for x on either side of 2.5 and calculate the corresponding y values. For example, let's choose x = 0 and x = 5:
For x = 0,
f(0) = 0^2 - 5(0) = 0
So we have the point (0,0).
For x = 5,
f(5) = 5^2 - 5(5) = 25 - 25 = 0
So we have the point (5,0).
Now we can plot these points on the graph and draw the parabolic curve passing through the points. The graph will be symmetric with respect to the axis of symmetry x = 2.5.
The graph should look similar to a "U" shape, opening upwards, and passing through the vertex (2.5,-6.25). The axis of symmetry is a vertical line passing through the vertex.
I hope this helps! Don't hesitate to ask if you have any further questions.