How do you identify the important parts of #f(x)= 2x^2 - 11# to graph it?

Question

Nathan Axel · Accepted Answer

Any quadratic equation of form #y=ax^2+bx+c# has a parabola graph.

If a is positive (like here a = 2) then the arms go up.

In this case, c = -11 and represents the y-intercept.

The roots or x-intercepts of the equation are when y = 0, So in this case when #2x^2-11=0 => x= + or -sqrt(11/2)#

The axis of symmetry is at #x=(-b)/(2a)=0# in this case.

Then you can combine it all to draw the graph.

graph{2x^2-11 [-25.66, 25.65, -12.83, 12.83]}

Eli Assouline · Answer

undefined To graph the function f(x) = 2x^2 - 11, you can follow these steps:
1. Identify the vertex: The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the point (h, k), where h = -b/(2a) and k = f(h). In this case, a = 2, b = 0, and c = -11. So, the x-coordinate of the vertex is h = -0/(2*2) = 0, and the y-coordinate is k = f(0) = 2(0)^2 - 11 = -11. Therefore, the vertex is (0, -11).
2. Determine the axis of symmetry: The axis of symmetry is the vertical line that passes through the vertex of the parabola. Since the vertex has an x-coordinate of 0, the axis of symmetry is the line x = 0, which is the y-axis.
3. Find the y-intercept: The y-intercept occurs when x = 0. So, plug x = 0 into the function: f(0) = 2(0)^2 - 11 = -11. Therefore, the y-intercept is (0, -11).
4. Plot additional points: You can choose other x-values and use them to find corresponding y-values to plot more points on the graph.
5. Draw the graph: Once you have plotted enough points, you can connect them smoothly to draw the graph of the function. Since the coefficient of x^2 is positive (2 in this case), the parabola opens upwards. Therefore, the graph will be a U-shaped curve that passes through the vertex (0, -11).